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Chapter 6 · 7 lessons

Linear Equations and Graphing

VI

CHAPTER 6 Linear Equations and Graphing

This graph illustrates the annual vehicle sales of gas motorcycles, gas cars, and electric vehicles from 1994 to 2010. It is a line graph with x– and y-axes, one of the most common types of graphs.

A line graph illustrating the annual vehicle sales of gas motorcycles, gas cars, and electric vehicles from 1994 to 2010. The x-axis ranges from 1994 through 2010 in two-year increments. The y-axis is labeled 0 to 30 million in increments of 5 millon per year. The y-axis is labeled “Annual Vehicle Sales (MM/year)” There are three line graphs. The first shows the annual sale of gas motorcycles from 5 million in 1994 to about 15 million in 2010. The next line is a green line labled EV for electric vehicles. It shows sales were null from 1994 through 2002, but they quickly rose to more than 25 million in sales per year. The last line is labeled gas cars and starts at 0 in 1994 and slowly rises from 2002 to 2010 to just over 10 million.
Image/graph source: Steve Jurvetson, Flickr.

Graphs are found in all areas of our lives—from commercials showing you which cell phone carrier provides the best coverage, to bank statements and news articles, to the boardroom of major corporations. In this chapter, we will study the rectangular coordinate system, which is the basis for most consumer graphs. We will look at linear graphs, slopes of lines, and equations of lines.

Attributions

This chapter has been adapted from the “Introduction” in Chapter 4 of Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

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6.1 Use the Rectangular Coordinate System

Learning Objectives

By the end of this section, you will be able to:

  • Plot points in a rectangular coordinate system
  • Verify solutions to an equation in two variables
  • Complete a table of solutions to a linear equation
  • Find solutions to a linear equation in two variables

Plot Points on a Rectangular Coordinate System

Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy-plane or the ‘coordinate plane.’

The horizontal number line is called the x-axis. The vertical number line is called the y-axis. The x-axis and the y-axis together form the rectangular coordinate system. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See (Figure 1).

‘Quadrant’ has the root ‘quad,’ which means ‘four.’

Described in previous paragraphs. Top right quadrant labelled “I”, top left “II”, bottom left “III”, and bottom right “IV”.
Figure .1

In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point.

Ordered pair

An ordered pair, (x,y),  gives the coordinates of a point in a rectangular coordinate system.Ordered pair x y. The first coordinate x labeled "x-coordinate", the second coordinate y labeled "y-coordinate".

The first number is the x-coordinate.

The second number is the y-coordinate.

The phrase ‘ordered pair’ means the order is important. What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is (0,0). The point (0,0) has a special name. It is called the origin.

The origin

The point (0,0) is called the origin. It is the point where the x-axis and y-axis intersect.

We use the coordinates to locate a point on the xy-plane. Let’s plot the point (1,3) as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through x=1. Then, locate 3 on the y-axis and sketch a horizontal line through y=3. Now, find the point where these two lines meet—that is the point with coordinates (1,3).

Figure 2. The result of the process described in previous paragraph plotting the point (1,3).
Figure .2

Notice that the vertical line through x=1 and the horizontal line through y=3 are not part of the graph. We just used them to help us locate the point (1,3).

EXAMPLE 1

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:

A (-5,4) B (-3,-4) C (2,-3) D (-2,3) E (3,5 over 2).

Solution

The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate.

  1. Since x=-5, the point is to the left of the y-axis. Also, since y=4, the point is above the x-axis. The point (-5,4) is in Quadrant II.
  2. Since x=-3, the point is to the left of the y-axis. Also, since y=-4, the point is below the x-axis. The point (-3,-4) is in Quadrant III.
  3. Since x=2, the point is to the right of the y-axis. Since y=-3, the point is below the x-axis. The point (2,-3) is in Quadrant lV.
  4. Since x=-2, the point is to the left of the y-axis. Since y=3, the point is above the x-axis. The point (-2,3) is in Quadrant II.
  5. Since x=3, the point is to the right of the y-axis. Since y=5 over 2, the point is above the x-axis. (It may be helpful to write 5 over 2 as a mixed number or decimal.) The point (3,5 over 2) is in Quadrant I.

A graph plotting the points (-5, 4), (-2, 3), (-3, -4), (3, 5/2), and (2, -3).

TRY IT 1.1

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:

A (-2,1) B (-3,-1) C (4,-4) D (-4,4) E (-4,3 over 2).

Show answer

                                                            A graph plotting the points described in the previous paragraph.

TRY IT 1.2

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:

A (-4,1) B (-2,3) C (2,-5) D (-2,5) E (-3,5 over 2)

Show answer

                                                            A graph plotting the points described in the previous paragraph.

How do the signs affect the location of the points? You may have noticed some patterns as you graphed the points in the previous example.

For the point in (Figure 2) in Quadrant IV, what do you notice about the signs of the coordinates? What about the signs of the coordinates of points in the third quadrant? The second quadrant? The first quadrant?

Can you tell just by looking at the coordinates in which quadrant the point (-2,5) is located? In which quadrant is (2,-5) located?

Quadrants

We can summarize sign patterns of the quadrants in this way.mathematical expression

                                            The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The top-right portion of the plane is labeled "I" and "ordered pair +, +", the top-left portion of the plane is labeled "II" and "ordered pair -, +", the bottom-left portion of the plane is labelled "III" "ordered pair -, -" and the bottom-right portion of the plane is labeled "IV" and "ordered pair +, -".

What if one coordinate is zero as shown in (Figure 3)? Where is the point (0,4) located? Where is the point (-2,0) located?                             

A graph plotting the points (0, 4) and (negative 2, 0).
Figure .3                                                

The point (0,4) is on the y-axis and the point (-2,0) is on the x-axis.

Points on the axes

Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates (a,0).

Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates (0,b).

EXAMPLE 2

Plot each point:A (0,5) B (4,0) C (-3,0) D (0,0) E (0,-1).

Solution

  1. Since x=0, the point whose coordinates are (0,5) is on the y-axis.
  2. Since y=0, the point whose coordinates are (4,0) is on the x-axis.
  3. Since y=0, the point whose coordinates are (-3,0) is on the x-axis.
  4. Since x=0 and y=0, the point whose coordinates are (0,0) is the origin.
  5. Since x=0, the point whose coordinates are (0,-1) is on the y-axis.
    A graph plotting the points (negative 3, 0), (0, 0), (0, negative 1), (0, 5), and (4, 0).

TRY IT 2.1

Plot each point: A (4,0) B (-2,0) C (0,0) D (0,2) E (0,-3).

Show answer

                                                            A graph plotting the points (4, 0), (negative 2, 0), (0, 0), (0, 2), and (0, negative 3).

TRY IT 2.2

Plot each point: A (-5,0) B (3,0) C (0,0) D (0,-1) E (0,4).

Show answer

                                                            A graph plotting the points (negative 5, 0), (3, 0), (0, 0), (0, negative 1), and (0, 4).

In algebra, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the x-coordinate of a point on a graph, read the number on the x-axis directly above or below the point. To identify the y-coordinate of a point, read the number on the y-axis directly to the left or right of the point. Remember, when you write the ordered pair use the correct order, (x,y).

EXAMPLE 3

Name the ordered pair of each point shown in the rectangular coordinate system.Described in following paragraph.

Solution

Point A is above -3 on the x-axis, so the x-coordinate of the point is -3.

  • The point is to the left of 3 on the y-axis, so the y-coordinate of the point is 3.
  • The coordinates of the point are (-3,3).

Point B is below -1 on the x-axis, so the x-coordinate of the point is -1.

  • The point is to the left of -3 on the y-axis, so the y-coordinate of the point is -3.
  • The coordinates of the point are (-1,-3).

Point C is above 2 on the x-axis, so the x-coordinate of the point is 2

  • The point is to the right of 4 on the y-axis, so the y-coordinate of the point is 4.
  • The coordinates of the point are (2,4).

Point D is below 4 on the x-axis, so the x-coordinate of the point is 4

  • The point is to the right of -4 on the y-axis, so the y-coordinate of the point is -4.
  • The coordinates of the point are (4,-4).

Point E is on the y-axis at y=-2. The coordinates of point E are (0,-2).

Point F is on the x-axis at x=3. The coordinates of point F are (3,0).

TRY IT 3.1

Name the ordered pair of each point shown in the rectangular coordinate system.                                                 A graph plotting the points (5, 1), (negative 2, 4), (negative 5, negative 1), (3, negative 2), (0, negative 5) labelled A-E.

Show answer

A: (5,1) B: (-2,4) C: (-5,-1) D: (3,-2) E: (0,-5) F: (4,0)

TRY IT 3.2

Name the ordered pair of each point shown in the rectangular coordinate system.                                                   A graph plotting the points (4, 2), (negative 2, 3), (negative 4, negative 4), (3, negative 5), (negative 3, 0) labelled A-E.

Show answer

A: (4,2) B: (-2,3) C: (-4,-4) D: (3,-5) E: (-3,0) F: (0,2)

Verify Solutions to an Equation in Two Variables

Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. The process of solving an equation ended with a statement like x=4. (Then, you checked the solution by substituting back into the equation.)
Here’s an example of an equation in one variable, and its one solution.

mathematical expression

But equations can have more than one variable. Equations with two variables may be of the form Ax+By=C. Equations of this form are called linear equations in two variables.

Linear equation

An equation of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.

Notice the word line in linear. Here is an example of a linear equation in two variables, x and y.

In this figure, we see the linear equation Ax plus By equals C. Below this is the equation x plus 4y equals 8. Below this are the values A equals 1, B equals 4, and C equals 8.

The equation y=-3x+5 is also a linear equation. But it does not appear to be in the form Ax+By=C. We can use the Addition Property of Equality and rewrite it in Ax+By=C form.

y=-3x+5
Add to both sides. y+3x=-3x+5+3x
Simplify. y+3x=5
Use the Commutative Property to put it in Ax+By=C form. 3x+y=5

By rewriting y=-3x+5 as 3x+y=5, we can easily see that it is a linear equation in two variables because it is of the form Ax+By=C. When an equation is in the form Ax+By=C, we say it is in standard form.

Standard Form of Linear Equation

A linear equation is in standard form when it is written Ax+By=C.

Most people prefer to have A, B, and C be integers and A greater than or equal to 0 when writing a linear equation in standard form, although it is not strictly necessary.

Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair (x,y). When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.

Solution of a Linear Equation in Two Variables

An ordered pair (x,y) is a solution of the linear equation Ax+By=C, if the equation is a true statement when the x– and y-values of the ordered pair are substituted into the equation.

EXAMPLE 4

Determine which ordered pairs are solutions to the equation x+4y=8.

A (0,2) B (2,-4) C (-4,3)

Solution

Substitute the x- and y-values from each ordered pair into the equation and determine if the result is a true statement.

This figure has three columns. At the top of the first column is the ordered pair (0, 2). Below this are the values x equals 0 and y equals 2. Below this is the equation x plus 4y equals 8. Below this is the same equation with 0 and 2 substituted for x and y: 0 plus 4 times 2 might equal 8. Below this is 0 plus 8 might equal 8. Below this is 8 equals 8 with a check mark next to it. Below this is the sentence “(0, 2) is a solution.” At the top of the second column is the ordered pair (2, negative 4). Below this are the values x equals 2 and y equals negative 4. Below this is the equation x plus 4y equals 8. Below this is the same equation with 2 and negative 4 substituted for x and y: 2 plus 4 times negative 4 might equal 8. Below this is 2 plus negative 16 might equal 8. Below this is negative 14 does not equal 8. Below this is the sentence: “(2, negative 4) is not a solution.” At the top of the third column is the ordered pair (negative 4, 3). Below this are the values x equals negative 4 and y equals 3. Below this is the equation x plus 4y equals 8. Below this is the same equation with negative 4 and 3 substituted for x and y: negative 4 plus 4 times 3 might equal 8. Below this is negative 4 plus 12 might equal 8. Below this is 8 equals 8 with a check mark next to it. Below this is the sentence: “(negative 4, 3) is a solution.”

TRY IT 4.1

Which of the following ordered pairs are solutions to 2x+3y=6?
A (3,0) B (2,0) C (6,-2)

Show answer

A, C

TRY IT 4.2

Which of the following ordered pairs are solutions to the equation 4x-y=8? A (0,8) B (2,0) C (1,-4)

Show answer

B, C

 

EXAMPLE 5

Which of the following ordered pairs are solutions to the equation y=5x-1?

A (0,-1) B (1,4) C (-2,-7)

Solution

Substitute the x– and y-values from each ordered pair into the equation and determine if it results in a true statement.

This figure has three columns. At the top of the first column is the ordered pair (0, negative 1). Below this are the values x equals 0 and y equals negative 1. Below this is the equation y equals 5x minus 1. Below this is the same equation with 0 and negative 1 substituted for x and y: negative 1 might equal 5 times 0 minus 1. Below this is negative 1 might equal 0 minus 1. Below this is negative 1 equals negative 1 with a check mark next to it. Below this is the sentence: “(0, negative 1) is a solution.” At the top of the second column is the ordered pair (1, 4). Below this are the values x equals 1 and y equals 4. Below this is the equation y equals 5x minus 1. Below this is the same equation with 1 and 4 substituted for x and y: 4 might equal 5 times 1 minus 1. Below this is 4 might equal 5 minus 1. Below this is 4 equals 4 with a check mark next to it. Below this is the sentence: “(1, 4) is a solution.” At the top of the right column is the ordered pair (negative 2, negative 7). Below this are the values x equals negative 2 and y equals negative 7. Below this is the equation y equals 5x minus 1. Below this is the same equation with negative 2 and negative 7 substituted for x and y: negative 7 might equal 5 times negative 2 minus 1. Below this is negative 7 might equal negative 10 minus 1. Below this is negative 7 does not equal negative 11. Below this is the sentence: “(negative 2, negative 7) is not a solution.”

TRY IT 5.1

Which of the following ordered pairs are solutions to the equation y=4x-3? A (0,3) B (1,1) C (-1,-1)

Show answer

B

TRY IT 5.2

Which of the following ordered pairs are solutions to the equation y=-2x+6? A (0,6) B (1,4) C (-2,-2)

Show answer

A, B

Complete a Table of Solutions to a Linear Equation in Two Variables

In the examples above, we substituted the x– and y-values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do you find the ordered pairs if they are not given? It’s easier than you might think—you can just pick a value for x and then solve the equation for y. Or, pick a value for y and then solve for x.

We’ll start by looking at the solutions to the equation y=5x-1 that we found in (Example 5). We can summarize this information in a table of solutions, as shown in (Table 1).

Table 1
y=5x-1
x y (x,y)
0 -1 (0,-1)
1 4 (1,4)

To find a third solution, we’ll let x=2 and solve for y.

                                                                  The figure shows the steps to solve for y when x equals 2 in the equation y equals 5 x minus 1. The equation y equals 5 x minus 1 is shown. Below it is the equation with 2 substituted in for x which is y equals 5 times 2 minus 1. To solve for y first multiply so that the equation becomes y equals 10 minus 1 then subtract so that the equation is y equals 9.

The ordered pair (2,9) is a solution to y=5x-1. We will add it to (Table 2).

Table 2
y=5x-1
x y (x,y)
0 -1 (0,-1)
1 4 (1,4)
2 9 (2,9)

We can find more solutions to the equation by substituting in any value of x or any value of y and solving the resulting equation to get another ordered pair that is a solution. There are infinitely many solutions of this equation.

EXAMPLE 6

Complete the table to find three solutions to the equation y=4x-2.

y=4x-2
x y (x,y)
0
-1
2

 

Solution

Substitute x=0, x=-1, and x=2 into y=4x-2.

This figure has three columns. At the top of the first column is the value x equals 0. Below this is the equation y equals 4x minus 2. Below this is the same equation with 0 substituted for x: y equals 4 times 0 minus 2. Below this is y equals 0 minus 2. Below this is y equals negative 2. Below this is the ordered pair (0, negative 2). At the top of the second column is the value x equals negative 1. Below this is the equation y equals 4x minus 2. Below this is the same equation with negative 1 substituted for x: y equals 4 times minus 1 minus 2. Below this is y equals negative 4 minus 2. Below this is y equals negative 6. Below this is the ordered pair (negative 1, negative 6). At the top of the third column is the value x equals 2. Below this is the equation y equals 4x minus 2. Below this is the same equation with 2 substituted for x: y equals 4 times 2 minus 2. Below this is y equals 8 minus 2. Below this is y equals 6. Below this is the ordered pair (2, 6).

The results are summarized in the table below.

y=4x-2
x y (x,y)
0 -2 (0,-2)
-1 -6 (-1,-6)
2   6 (2,6)

TRY IT 6.1

Complete the table to find three solutions to this equation: y=3x-1.

y=3x-1
x y (x,y)
0
-1
2
Show answer
y=3x-1
x y (x,y)
0 -1 (0,-1)
-1 -4 (-1,-4)
2 5 (2,5)

TRY IT 6.2

Complete the table to find three solutions to this equation: y=6x+1.

y=6x+1
x y (x,y)
0
1
-2

 

Show answer
y=6x+1
x y (x,y)
0 1 (0,1)
1 7 (1,7)
-2 -11 (-2,-11)

EXAMPLE 7

Complete the table to find three solutions to the equation 5x-4y=20.

5x-4y=20
x y (x,y)
0
0
5
Solution

Substitute the given value into the equation 5x-4y=20 and solve for the other variable. Then, fill in the values in the table.

This figure has three columns. At the top of the first column is the value x equals 0. Below this is the equation 5x minus 4y equals 20. Below this is the same equation with 0 substituted for x: 5 times 0 minus 4y equals 20. Below this is 0 minus 4y equals 20. Below this is negative 4y equals 20. Below this is y equals negative 5. Below this is the ordered pair (0, negative 5). At the top of the second column is the value y equals 0. Below this is the equation 5x minus 4y equals 20. Below this is the same equation with 0 substituted for y: 5x minus 4 times 0 equals 20. Below this is 5x minus 0 equals 20. Below this is 5x equals 20. Below this is x equals 4. Below this is the ordered pair (4, 0). At the top of the third column is the value y equals 5. Below this is the equation 5x minus 47 equals 20. Below this is the same equation with 5 substituted for y: 5x minus 4 times 5 equals 20. Below this is the equation 5x minus 20 equals 20. Below this is 5x equals 40. Below this is x equals 8. Below this is the ordered pair (8, 5).

The results are summarized in the table below.

5x-4y=20
x y (x,y)
0 -5 (0,-5)
4 0 (4,0)
8 5 (8,5)

TRY IT 7.1

Complete the table to find three solutions to this equation: 2x-5y=20.

2x-5y=20
x y (x,y)
0
0
-5
Show answer
2x-5y=20
x y (x,y)
0 -4 (0,-4)
10 0 (10,0)
-5 -6 (-5,-6)

TRY IT 7.2

Complete the table to find three solutions to this equation: 3x-4y=12.

3x-4y=12
x y (x,y)
0
0
-4
Show answer
3x-4y=12
x y (x,y)
0 -3 (0,-3)
4 0 (4,0)
-4 -6 (-4,-6)

Find Solutions to a Linear Equation

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for x or y. But since you’ll need to use that number to solve for the other variable it’s a good idea to choose a number that’s easy to work with.

When the equation is in y-form, with the y by itself on one side of the equation, it is usually easier to choose values of x and then solve for y.

EXAMPLE 8

Find three solutions to the equation y=-3x+2.

Solution

We can substitute any value we want for x or any value for y. Since the equation is in y-form, it will be easier to substitute in values of x. Let’s pick x=0, x=1, and x=-1.

. . .
Substitute the value into the equation. . . .
Simplify. . . .
Simplify. . . .
Write the ordered pair. . . .
Check. (0, 2) (1, −1) (−1, 5)
y=-3x+2 y=-3x+2 y=-3x+2
2=-3 times 0+2 -1=-3 times 1+2 5=-3(-1)+2
2=0+2 -1=-3+2 5=3+2
mathematical expression mathematical expression mathematical expression

So, (0,2), (1,-1) and (-1,5) are all solutions to y=-3x+2. We show them in table below.

y=-3x+2
x y (x,y)
0 2 (0,2)
1 -1 (1,-1)
-1 5 (-1,5)

TRY IT 8.1

Find three solutions to this equation: y=-2x+3.

Show answer

Answers will vary.

TRY IT 8.2

Find three solutions to this equation: y=-4x+1.

Show answer

Answers will vary

We have seen how using zero as one value of x makes finding the value of y easy. When an equation is in standard form, with both the x and y on the same side of the equation, it is usually easier to first find one solution when x=0 find a second solution when y=0, and then find a third solution.

EXAMPLE 9

Find three solutions to the equation 3x+2y=6.

Solution

We can substitute any value we want for x or any value for y. Since the equation is in standard form, let’s pick first x=0, then y=0, and then find a third point.

. . .
. . .
Substitute the value into the equation. . . .
Simplify. . . .
Solve. . . .
. . .
Write the ordered pair. (0, 3) (2, 0) (1,3 over 2)
Check. 3x+2y=6 3x+2y=6 3x+2y=6
3 times 0+2 times 3=6 3 times 2+2 times 0=6 3 times 1+2 times 3 over 2=6
0+6=6 6+0=6 3+3=6
mathematical expression 6=6 6=6

So (0,3), (2,0), and (1,3 over 2) are all solutions to the equation 3x+2y=6. We can list these three solutions in the table below.

3x+2y=6
x y (x,y)
0 3 (0,3)
2 0 (2,0)
1 3 over 2 (1,3 over 2)

EXAMPLE 9.1

Find three solutions to the equation 2x+3y=6.

Show answer

Answers will vary.

TRY IT 9.2

Find three solutions to the equation 4x+2y=8.

Show answer

Answers will vary.

Glossary

linear equation
A linear equation is of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.
ordered pair
An ordered pair (x,y) gives the coordinates of a point in a rectangular coordinate system.
origin
The point (0,0) is called the origin. It is the point where the x-axis and y-axis intersect.
quadrant
The x-axis and the y-axis divide a plane into four regions, called quadrants.
rectangular coordinate system
A grid system is used in algebra to show a relationship between two variables; also called the xy-plane or the ‘coordinate plane.’
x-coordinate
The first number in an ordered pair (x,y).
y-coordinate
The second number in an ordered pair (x,y).

Practice Makes Perfect

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

1.A (-4,2)
B (-1,-2)
C (3,-5)
D (-3,5)
E (5 over 3,2)

2. A (-2,-3)
B (3,-3)
C (-4,1)
D (4,-1)
E (3 over 2,1)
3. A (3,-1)
B (-3,1)
C (-2,2)
D (-4,-3)
E (1,14 over 5)
4. A (-1,1)
B (-2,-1)
C (2,1)
D (1,-4)
E (3,7 over 2)

In the following exercises, plot each point in a rectangular coordinate system.

5. A (-2,0)
B (-3,0)
C (0,0)
D (0,4)
E (0,2)
6. A (0,1)
B (0,-4)
C (-1,0)
D (0,0)
E (5,0)
7. A (0,0)
B (0,-3)
C (-4,0)
D (1,0)
E (0,-2)
8. A (-3,0)
B (0,5)
C (0,-2)
D (2,0)
E (0,0)

In the following exercises, name the ordered pair of each point shown in the rectangular coordinate system.

9.
A graph plotting the points A (negative 4, 1), B (negative 3, negative 4), C (1, negative 3), D (4, 3).

10.

A graph plotting the points A (negative 4, 2), B (3, 5), C (negative 4, negative 2), D (5, negative 1).
11.
A graph plotting the points A (0, negative 2), B (negative 2, 0), C (0, 5), D (5, 0).

12.

A graph plotting the points A (0, negative 1), B (negative 1, 0), C (4, 0), D (0, 4).

Verify Solutions to an Equation in Two Variables

In the following exercises, which ordered pairs are solutions to the given equations?

13. 2x+y=6

A (1,4)
B (3,0)
C (2,3)

14. x+3y=9

A (0,3)
B (6,1)
C (-3,-3)

15. 4x-2y=8

A (3,2)
B (1,4)
C (0,-4)

16. 3x-2y=12

A (4,0)
B (2,-3)
C (1,6)

17. y=4x+3

A (4,3)
B (-1,-1)
C (1 over 2,5)

18. y=2x-5

A (0,-5)
B (2,1)
C (1 over 2,-4)

19. y=1 over 2x-1

A (2,0)
B (-6,-4)
C (-4,-1)

20. y=1 over 3x+1

A (-3,0)
B (9,4)
C (-6,-1)

Complete a Table of Solutions to a Linear Equation

In the following exercises, complete the table to find solutions to each linear equation.

21. y=2x-4
x y (x,y)
0
2
-1
22. y=3x-1
x y (x,y)
0
2
-1

 23. y=-x+5

x y (x,y)
0
3
-2
 24. y=-x+2
x y (x,y)
0
3
-2
 25. y=1 over 3x+1
x y (x,y)
0
3
6

 26. y=1 over 2x+4

x y (x,y)
0
2
4

 27. y=-3 over 2x-2

x y (x,y)
0
2
-2

 28. y=-2 over 3x-1

x y (x,y)
0
3
-3

29. x+3y=6

x y (x,y)
0
3
0

  30. x+2y=8

x y (x,y)
0
4
0

31. 2x-5y=10

x y (x,y)
0
10
0

32. 3x-4y=12

x y (x,y)
0
8
0

Find Solutions to a Linear Equation

In the following exercises, find three solutions to each linear equation.

33. y=5x-8 34. y=3x-9
35. y=-4x+5 36. y=-2x+7
37. x+y=8 38. x+y=6
39. x+y=-2 40. x+y=-1
41. 3x+y=5 42. 2x+y=3
43. 4x-y=8 44. 5x-y=10
45. 2x+4y=8 46. 3x+2y=6
47. 5x-2y=10 48. 4x-3y=12

Everyday Math

49. Weight of a baby. Mackenzie recorded her baby’s weight every two months. The baby’s age, in months, and weight, in pounds, are listed in the table below, and shown as an ordered pair in the third column.

a) Plot the points on a coordinate plane.

The x y axis with no points plotted.

b) Why is only Quadrant I needed?

Age x Weight y (x,y)
0 7 (0, 7)
2 11 (2, 11)
4 15 (4, 15)
6 16 (6, 16)
8 19 (8, 19)
10 20 (10, 20)
12 21 (12, 21)

50. Weight of a child. Latresha recorded her son’s height and weight every year. His height, in inches, and weight, in pounds, are listed in the table below, and shown as an ordered pair in the third column.

a) Plot the points on a coordinate plane.

The x y axis with no points plotted.

b) Why is only Quadrant I needed?

Height x Weight y (x,y)
28 22 (28, 22)
31 27 (31, 27)
33 33 (33, 33)
37 35 (37, 35)
40 41 (40, 41)
42 45 (42, 45)

Writing Exercises

51. Explain in words how you plot the point (4,-2) in a rectangular coordinate system. 52. How do you determine if an ordered pair is a solution to a given equation?
53. Is the point (-3,0) on the x-axis or y-axis? How do you know? 54. Is the point (0,8) on the x-axis or y-axis? How do you know?

Answers

1.

A graph plotting the points a (negative 4, 2), b (negative 1, negative 2), c (3, negative 5), d (negative 3, 5), e (5 thirds, 2).  

3.

A graph plotting the points a (3, negative 1), b (negative 3, 1), c (negative 2, 2), d (negative 4, negative 3), e (1, 14 fifths).

5.

A graph plotting the points a (negative 2, 0), b (negative 3, 0), c (0, 0), d (0, 4), e (0, 3).

7.

A graph plotting the points a (0, 0), b (0, negative 3), c (negative 4, 0), d (1, 0), e (0, negative 2).

9. A: (-4,1) B: (-3,-4) C: (1,-3) D: (4,3) 11. A: (0,-2) B: (-2,0) C: (0,5) D: (5,0)
13. A, B 15. A, C
17. B, C 19. A, B
21.
x y (x,y)
0 -4 (0,-4)
2 0 (2,0)
-1 -6 (-1,-6)
23.
x y (x,y)
0 5 (0,5)
3 2 (3,2)
-2 7 (-2,7)
25.
x y (x,y)
0 1 (0,1)
3 2 (3,2)
6 3 (6,3)
25.
x y (x,y)
0 1 (0,1)
3 2 (3,2)
6 3 (6,3)
27.
x y (x,y)
0 -2 (0,-2)
2 -5 (2,-5)
-2 1 (-2,1)
29.
x y (x,y)
0 2 (0,2)
3 4 (3,1)
6 0 (6,0)
31.
x y (x,y)
0 -2 (0,-2)
10 2 (10,2)
5 0 (5,0)
33. Answers will vary.
35. Answers will vary. 37. Answers will vary.
39. Answers will vary. 41. Answers will vary.
43. Answers will vary. 45. Answers will vary.
47. Answers will vary. 49.

a)

A graph that plots the points (0, 7), (2, 11), (4, 15), (6, 16), (8, 19), (10, 20) and (12, 21).

b) Age and weight are only positive.

51. Answers will vary. 53. Answers will vary.

Attributions

This chapter has been adapted from “Use the Rectangular Coordinate System” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

33

6.2 Graph Linear Equations in Two Variables

Learning Objectives

By the end of this section, you will be able to:

  • Recognize the relationship between the solutions of an equation and its graph.
  • Graph a linear equation by plotting points.
  • Graph vertical and horizontal lines.

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the previous section, we found several solutions to the equation 3x+2y=6. They are listed in the table below. So, the ordered pairs (0,3), (2,0), and (1,3 over 2) are some solutions to the equation 3x+2y=6. We can plot these solutions in the rectangular coordinate system as shown in (Figure 1).

3x+2y=6
x y (x,y)
0 3 (0,3)
2 0 (2,0)
1 3 over 2 (1,3 over 2)
A graph that plots the points (0, 3), (1, three halves), and (2, 0).
Figure .1

Notice how the points line up perfectly? We connect the points with a line to get the graph of the equation 3x+2y=6. See (Figure 2). Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

Described in previous paragraph.
Figure .2

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions.

Notice that the point whose coordinates are (-2,6) is on the line shown in (Figure 3). If you substitute x=-2 and y=6 into the equation, you find that it is a solution to the equation.

Graphs the equation 3x plus 2y equals 6. The points (negative 2, 6) and (4, 1) are plotted. The line goes through (−2, 6) but not (4, 1).
Figure .3

The figure shows a series of equations to check if the ordered pair (negative 2, 6) is a solution to the equation 3x plus 2y equals 6. The first line states “Test (negative 2, 6)”. The negative 2 is colored blue and the 6 is colored red. The second line states the two- variable equation 3x plus 2y equals 6. The third line shows the ordered pair substituted into the two- variable equation resulting in 3(negative 2) plus 2(6) equals 6 where the negative 2 is colored blue to show it is the first component in the ordered pair and the 6 is red to show it is the second component in the ordered pair. The fourth line is the simplified equation negative 6 plus 12 equals 6. The fifth line is the further simplified equation 6equals6. A check mark is written next to the last equation to indicate it is a true statement and show that (negative 2, 6) is a solution to the equation 3x plus 2y equals 6.

So the point (-2,6) is a solution to the equation 3x+2y=6. (The phrase “the point whose coordinates are (-2,6)” is often shortened to “the point (-2,6).”)

The figure shows a series of equations to check if the ordered pair (4, 1) is a solution to the equation 3x plus 2y equals 6. The first line states “What about (4, 1)?”. The 4 is colored blue and the 1 is colored red. The second line states the two- variable equation 3x plus 2y equals 6. The third line shows the ordered pair substituted into the two- variable equation resulting in 3(4) plus 2(1) equals 6 where the 4 is colored blue to show it is the first component in the ordered pair and the 1 is red to show it is the second component in the ordered pair. The fourth line is the simplified equation 12 plus 2 equals 6. A question mark is placed above the equals sign to indicate that it is not known if the equation is true or false. The fifth line is the further simplified statement 14 not equal to 6. A “not equals” sign is written between the two numbers and looks like an equals sign with a forward slash through it.

So (4,1) is not a solution to the equation 3x+2y=6. Therefore, the point (4,1) is not on the line. See (Figure 2). This is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to the equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation 3x+2y=6.

Graph of a linear equation

The graph of a linear equation Ax+By=C is a line.

  • Every point on the line is a solution of the equation.
  • Every solution of this equation is a point on this line.

EXAMPLE 1

The graph of y=2x-3 is shown.

Graphs the line 2x−3.

For each ordered pair, decide:

a) Is the ordered pair a solution to the equation?
b) Is the point on the line?

A (0,-3) B (3,3) C (2,-3) D (-1,-5)

Solution

Substitute the x– and y– values into the equation to check if the ordered pair is a solution to the equation.

  1. The figure shows a series of equations to check if the ordered pairs (0, negative 3), (3, 3), (2, negative 3), and (negative 1, negative 5) are a solutions to the equation y equals 2x negative 3. The first line states the ordered pairs with the labels A: (0, negative 3), B: (3, 3), C: (2, negative 3), and D: (negative 1, negative 5). The first components are colored blue and the second components are colored red. The second line states the two- variable equation y equals 2x minus 3. The third line shows the four ordered pairs substituted into the two- variable equation resulting in four equations. The first equation is negative 3 equals 2(0) minus 3 where the 0 is colored clue and the negative 3 on the left side of the equation is colored red. The second equation is 3 equals 2(3) minus 3 where the 3 in parentheses is colored clue and the 3 on the left side of the equation is colored red. The third equation is negative 3 equals 2(2) minus 3 where the 2 in parentheses is colored clue and the negative 3 on the left side of the equation is colored red. The fourth equation is negative 5 equals 2(negative 1) minus 3 where the negative 1 is colored clue and the negative 5 is colored red. Question marks are placed above all the equal signs to indicate that it is not known if the equations are true or false. The fourth line shows the simplified versions of the four equations. The first is negative 3 equals negative 3 with a check mark indicating (0, negative 3) is a solution. The second is 3 equals 3 with a check mark indicating (3, 3) is a solution. The third is negative 3 not equals 1 indicating (2, negative 3) is not a solution. The fourth is negative 5 equals negative 5 with a check mark indicating (negative 1, negative 5) is a solution.
  2. Plot the points A (0,3), B (3,3), C (2,-3), and D (-1,-5).
    Graph of the equation 2x−3. The points described in the previous paragraph are plotted.

The points (0,3), (3,3), and (-1,-5) are on the line y=2x-3, and the point (2,-3) is not on the line.

The points that are solutions to y=2x-3 are on the line, but the point that is not a solution is not on the line.

TRY IT 1.1

Use the graph of y=3x-1 to decide whether each ordered pair is:

  • a solution to the equation.
  • on the line.

a) (0,-1) b) (2,5)

Graph of the equation y = 3x−1.

Show answer

a) yes, yes b) yes, yes

TRY IT 1.2

Use graph of y=3x-1 to decide whether each ordered pair is:

  • a solution to the equation
  • on the line

a) (3,-1) b) (-1,-4)

Graph of the equation y = 3x−1.

Show answer

a) no, no b) yes, yes

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used to graph 3x+2y=6 is called plotting points, or the Point–Plotting Method.

EXAMPLE 2

How To Graph an Equation By Plotting Points

Graph the equation y=2x+1 by plotting points.

Solution

The figure shows the three step procedure for graphing a line from the equation using the example equation y equals 2x minus 1. The first step is to “Find three points whose coordinates are solutions to the equation. Organize the solutions in a table”. The remark is made that “You can choose any values for x or y. In this case, since y is isolated on the left side of the equation, it is easier to choose values for x”. The work for the first step of the example is shown through a series of equations aligned vertically. From the top down, the equations are y equals 2x plus 1, x equals 0 (where the 0 is blue), y equals 2x plus 1, y equals 2(0) plus 1 (where the 0 is blue), y equals 0 plus 1, y equals 1, x equals 1 (where the 1 is blue), y equals 2x plus 1, y equals 2(1) plus 1 (where the 1 is blue), y equals 2 plus 1, y equals 3, x equals negative 2 (where the negative 2 is blue), y equals 2x plus 1, y equals 2(negative 2) plus 1 (where the negative 2 is blue), y equals negative 4 plus 1, y equals negative 3. The work is then organized in a table. The table has 5 rows and 3 columns. The first row is a title row with the equation y equals 2x plus 1. The second row is a header row and it labels each column. The first column header is “x”, the second is “y” and the third is “(x, y)”. Under the first column are the numbers 0, 1, and negative 2. Under the second column are the numbers 1, 3, and negative 3. Under the third column are the ordered pairs (0, 1), (1, 3), and (negative 2, negative 3).The second step is to “Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!” For the example the points are (0, 1), (1, 3), and (negative 2, negative 3). A graph shows the three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). The question “Do the points line up?” is stated and followed with the answer “Yes, the points line up.”The third step of the procedure is “Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.” A graph shows a straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). A straight line goes through all three points. The line has arrows on both ends pointing to the edge of the figure. The line is labeled with the equation y equals 2x plus 1. The statement “This line is the graph of y equals 2x plus 1” is included next to the graph.

TRY IT 2.1

Graph the equation by plotting points: y=2x-3.

Show answer
Graph of the equation y = 2x−3.

TRY IT 2.2

Graph the equation by plotting points: y=-2x+4.

Show answer
Graph of the equation y = −2+4.

HOW TO: Graph a linear equation by plotting points.

The steps to take when graphing a linear equation by plotting points are summarized below.

  1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
  2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
  3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line.

If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. Look at the difference between part (a) and part (b) in (Figure 4).

Figure a shows three points with a straight line through them. Figure b shows three points that do not lie on the same line.
Figure .4

Let’s do another example. This time, we’ll show the last two steps all on one grid.

EXAMPLE 3

Graph the equation y=-3x.

Solution

Find three points that are solutions to the equation. Here, again, it’s easier to choose values for x. Do you see why?

The figure shows three sets of equations used to determine ordered pairs from the equation y equals negative 3x. The first set has the equations: x equals 0 (where the 0 is blue), y equals negative 3x, y equals negative 3(0) (where the 0 is blue), y equals 0. The second set has the equations: x equals 1 (where the 1 is blue), y equals negative 3x, y equals negative 3(1) (where the 1 is blue), y equals negative 3. The third set has the equations: x equals negative 2 (where the negative 2 is blue), y equals negative 3x, y equals negative 3(negative 2) (where the negative 2 is blue), y equals 6.

We list the points in the table below.

y=-3x
x y (x,y)
0 0 (0,0)
1 -3 (1,-3)
-2 6 (-2,6)

Plot the points, check that they line up, and draw the line.

Graph of the equation y = −3x. The points listed in the previous table are plotted.

 

TRY IT 3.1

Graph the equation by plotting points: y=-4x.

Show answer
A graph of the equation y = −4x.

EXAMPLE 3.2

Graph the equation by plotting points: y=x.

Show answer
A graph of the equation y = x.

When an equation includes a fraction as the coefficient of x, we can still substitute any numbers for x. But the math is easier if we make ‘good’ choices for the values of x. This way we will avoid fraction answers, which are hard to graph precisely.

EXAMPLE 4

Graph the equation y=1 over 2x+3.

Solution

Find three points that are solutions to the equation. Since this equation has the fraction 1 over 2 as a coefficient of x, we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of 2 a good choice for values of x?

The figure shows three sets of equations used to determine ordered pairs from the equation y equals (one half)x plus 3. The first set has the equations: x equals 0 (where the 0 is blue), y equals (one half)x plus 3, y equals (one half)(0) plus 3 (where the 0 is blue), y equals 0 plus 3, y equals 3. The second set has the equations: x equals 2 (where the 2 is blue), y equals (one half)x plus 3, y equals (one half)(2) plus 3 (where the 2 is blue), y equals 1 plus 3, y equals 4. The third set has the equations: x equals 4 (where the 4 is blue), y equals (one half)x plus 3, y equals (one half)(4) plus 3 (where the 4 is blue), y equals 2 plus 3, y equals 5.

The points are shown in the table below.

y=1 over 2x+3
x y (x,y)
0 3 (0,3)
2 4 (2,4)
4 5 (4,5)

Plot the points, check that they line up, and draw the line.

The points listed in the previous table are plotted. The equation y = 1 half x + 3 is graphed.

TRY IT 4. 1

Graph the equation y=1 over 3x-1.

Show answer
A graph of the equation y = 1 third x−1.

TRY IT 4.2

Graph the equation y=1 over 4x+2.

Show answer
A graph of the equation y = 1 fourth + 2.

So far, all the equations we graphed had y given in terms of x. Now we’ll graph an equation with x and y on the same side. Let’s see what happens in the equation 2x+y=3. If y=0 what is the value of x?

The figure shows a set of equations used to determine an ordered pair from the equation 2x plus y equals 3. The first equation is y equals 0 (where the 0 is red). The second equation is the two- variable equation 2x plus y equals 3. The third equation is the onenegative variable equation 2x plus 0 equals 3 (where the 0 is red). The fourth equation is 2x equals 3. The fifth equation is x equals three halves. The last line is the ordered pair (three halves, 0).

This point has a fraction for the x– coordinate and, while we could graph this point, it is hard to be precise graphing fractions. Remember in the example y=1 over 2x+3, we carefully chose values for x so as not to graph fractions at all. If we solve the equation 2x+y=3 for y, it will be easier to find three solutions to the equation.

mathematical expression

The solutions for x=0, x=1, and x=-1 are shown in the table below. The graph is shown in (Figure 5).

2x+y=3
x y (x,y)
0 3 (0,3)
1 1 (1,1)
-1 5 (-1,5)

 

The points listed in the previous table are plotted. The equation 2x + y = 3 is graphed.
Figure .5

Can you locate the point (3 over 2,0), which we found by letting y=0, on the line?

EXAMPLE 5

Graph the equation 3x+y=-1.

Solution
Find three points that are solutions to the equation. mathematical expression
First, solve the equation for y. mathematical expression

We’ll let x be 0, 1, and -1 to find 3 points. The ordered pairs are shown in the table below. Plot the points, check that they line up, and draw the line. See (Figure 6).

3x+y=-1
x y (x,y)
0 -1 (0,-1)
1 -4 (1,-4)
-1 2 (-1,2)
The points listed in the previous table are plotted. The equation 3x+y = −1 is graphed.
Figure .6

EXAMPLE 5.1

Graph the equation 2x+y=2.

Show answer
Graph of the equation 2 x + y = 2.

TRY IT 5.2

Graph the equation 4x+y=-3.

Show answer
Graph of the equation 4 x + y = −3.

If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the x– and y-axis are the same, the graphs match!

The equation in (Example 5) was written in standard form, with both x and y on the same side. We solved that equation for y in just one step. But for other equations in standard form it is not that easy to solve for y, so we will leave them in standard form. We can still find a first point to plot by letting x=0 and solving for y. We can plot a second point by letting y=0 and then solving for x. Then we will plot a third point by using some other value for x or y.

EXAMPLE 6

Graph the equation 2x-3y=6.

Solution
Find three points that are solutions to the equation. mathematical expression
First, let x=0. mathematical expression
Solve for y. mathematical expression
Now let y=0. mathematical expression
Solve for x. mathematical expression
We need a third point. Remember, we can choose any value for x or y. We’ll let x=6. mathematical expression
Solve for y. mathematical expression

We list the ordered pairs in the table below. Plot the points, check that they line up, and draw the line. See (Figure 7).

2x-3y=6
x y (x,y)
0 -2 (0,-2)
3 0 (3,0)
6 2 (6,2)
The points listed in previous table are plotted. The equation 2x − 3y = 6 is plotted.
Figure .7

TRY IT 6.1

Graph the equation 4x+2y=8.

Show answer
Graph of the equation 4x + 2y = 8.

TRY IT 6.2

Graph the equation 2x-4y=8.

Show answer
Graph of the equation 2x − 3y = 8.

Graph Vertical and Horizontal Lines

Can we graph an equation with only one variable? Just x and no y, or just y without an x? How will we make a table of values to get the points to plot?

Let’s consider the equation x=-3. This equation has only one variable, x. The equation says that x is always equal to -3, so its value does not depend on y. No matter what y is, the value of x is always -3.

So to make a table of values, write -3 in for all the x values. Then choose any values for y. Since x does not depend on y, you can choose any numbers you like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the y-coordinates. See the table below.

x=-3
x y (x,y)
-3 1 (-3,1)
-3 2 (-3,2)
-3 3 (-3,3)

Plot the points from the table and connect them with a straight line. Notice in (Figure 8) that we have graphed a vertical line.

The points listed in the previous table are plotted. The equation x = −3 is graphed. The resulting line is vertical.
Figure .8

Vertical line

A vertical line is the graph of an equation of the form x=a.

The line passes through the x-axis at (a,0).

EXAMPLE 7

Graph the equation x=2.

Solution

The equation has only one variable, x, and x is always equal to 2. We create the table below where x is always 2 and then put in any values for y. The graph is a vertical line passing through the x-axis at 2. See (Figure 9).

x=2
x y (x,y)
2 1 (2,1)
2 2 (2,2)
2 3 (2,3)
The points listed in the previous table are plotted. The equation x = 2 is graphed. The resulting line is vertical.
Figure .9

TRY IT 7.1

Graph the equation x=5.

Show answer
Graph of the equation x = 5. The resulting line is vertical.

TRY IT 7.2

Graph the equation x=-2.

Show answer
Graph of the equation x = −2. The resulting line is vertical.

What if the equation has y but no x? Let’s graph the equation y=4. This time the y– value is a constant, so in this equation, y does not depend on x. Fill in 4 for all the y’s in the table below and then choose any values for x. We’ll use 0, 2, and 4 for the x-coordinates.

y=4
x y (x,y)
0 4 (0,4)
2 4 (2,4)
4 4 (4,4)

The graph is a horizontal line passing through the y-axis at 4. See (Figure 10).

The points listed in the previous table are plotted. The equation y = 4 is graphed. The resulting line is horizontal.
Figure .10

Horizontal line

A horizontal line is the graph of an equation of the form y=b.

The line passes through the y-axis at (0,b).

EXAMPLE 8

Graph the equation y=-1.

Solution

The equation y=-1 has only one variable, y. The value of y is constant. All the ordered pairs in the table below have the same y-coordinate. The graph is a horizontal line passing through the y-axis at -1, as shown in (Figure 11).

y=-1
x y (x,y)
0 -1 (0,-1)
3 -1 (3,-1)
-3 -1 (-3,-1)
The points listed in the previous table are plotted. The equation y = −1 is graphed. The resulting line is horizontal.
Figure .11

TRY IT 8.1

Graph the equation y=-4.

Show answer
Graph of the equation y = −4. The resulting line is horizontal.

TRY IT 8.2

Graph the equation y=3.

Show answer
Graph of the equation y = 3. The resulting line is horizontal.

The equations for vertical and horizontal lines look very similar to equations like y=4x. What is the difference between the equations y=4x and y=4?

The equation y=4x has both x and y. The value of y depends on the value of x. The y-coordinate changes according to the value of x. The equation y=4 has only one variable. The value of y is constant. The y-coordinate is always 4. It does not depend on the value of x. See the table below.

y=4x y=4
x y (x,y) x y (x,y)
0 0 (0,0) 0 4 (0,4)
1 4 (1,4) 1 4 (1,4)
2 8 (2,8) 2 4 (2,4)
The equations y = 4 and y = 4x are graphed and labelled.
Figure .12

Notice, in (Figure 12), the equation y=4x gives a slanted line, while y=4 gives a horizontal line.

EXAMPLE 9

Graph y=-3x and y=-3 in the same rectangular coordinate system.

Solution

Notice that the first equation has the variable x, while the second does not. See the table below. The two graphs are shown in (Figure 13).

y=-3x y=-3
x y (x,y) x y (x,y)
0 0 (0,0) 0 -3 (0,-3)
1 -3 (1,-3) 1 -3 (1,-3)
2 -6 (2,-6) 2 -3 (2,-3)
The equations y = −3 and y = −3x are graphed and labelled. The equation y = −3x is a slanted line while y = −3 is horizontal.
Figure .13

TRY IT 9.1

Graph y=-4x and y=-4 in the same rectangular coordinate system.

Show answer
The equations y = −4 and y = −4x are graphed and labelled. The equation y = −4x is a slanted line while y = −4 is horizontal.

TRY IT 9.2

Graph y=3 and y=3x in the same rectangular coordinate system.

Show answer
The equations y = 3 and y = 3x are graphed and labelled. The equation y = 3x is a slanted line while y = 3 is horizontal.

Key Concepts

  • Graph a Linear Equation by Plotting Points
    1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
    2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!
    3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

Glossary

graph of a linear equation
The graph of a linear equation Ax+By=C is a straight line. Every point on the line is a solution of the equation. Every solution of this equation is a point on this line.
horizontal line
A horizontal line is the graph of an equation of the form y=b. The line passes through the y-axis at (0,b).
vertical line
A vertical line is the graph of an equation of the form x=a. The line passes through the x-axis at (a,0).

Practice Makes Perfect

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

a) Is the ordered pair a solution to the equation? b) Is the point on the line?

1. y=x+2

a) (0,2)
b) (1,2)
c) (-1,1)
d) (-3,-1)

Graph of the equation y = x + 2.

2. y=x-4

a) (0,-4)
b) (3,-1)
c) (2,2)
d) (1,-5)

Graph of the equation y = x − 4.

3. y=1 over 2x-3

a) (0,-3)
b) (2,-2)
c) (-2,-4)
d) (4,1)

Graph of the equation y = 1 half x − 3.

4. y=1 over 3x+2

a) (0,2)
b) (3,3)
c) (-3,2)
d) (-6,0)

Graph of the equation y = 1 third x + 2.

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

5. y=3x-1 6. y=2x+3
7. y=-3x+3 8. y=-3x+1
9. y=x+2 10. y=x-3
11. y=-x-3 12. y=-x-2
13. y=2x 14. y=3x
15. y=3x 16. y=-2x
17. y=1 over 2x+2 18. y=1 over 3x-1
19. y=4 over 3x-5 20. y=3 over 2x-3
21. y=-2 over 5x+1 22. y=-4 over 5x-1
23. y=-3 over 2x+2 24. y=-5 over 3x+4
25. x+y=6 26. x+y=4
27. x+y=-3 28. x+y=-3
29. x-y=2 30. x-y=1
31. x-y=-1 32. x-y=-3
33. 3x+y=7 34. 5x+y=6
35. 2x+y=-3 36. 4x+y=-5
37. 1 over 3x+y=2 38. 1 over 2x+y=3
39. 2 over 5x+y=-4 40. 3 over 4x-y=6
41. 2x+3y=12 42. 4x+2y=12
43. 3x-4y=12 44. 2x-5y=10
45. x-6y=3 46. x-4y=2
47. 3x+y=2 48. 3x+5y=5

Graph Vertical and Horizontal Lines

In the following exercises, graph each equation.

49. x=4 50. x=3
51. x=-2 52. x=-5
53. y=3 54. y=1
55. y=-5 56. y=-2
57. x=7 over 3 58. x=5 over 4
59. y=-15 over 4 60. y=-5 over 3

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

61. y=2x and y=2 62. y=5x and y=5
63. y=-1 over 2x and y=-1 over 2 64. y=-1 over 3x and y=-1 over 3

Mixed Practice

In the following exercises, graph each equation.

65. y=4x 66. y=2x
67. y=-1 over 2x+3 68. y=1 over 4x-2
69. y=-x 70. y=x
71. x-y=3 72. x+y=-5
73. 4x+y=2 74. 2x+y=6
75. y=-1 76. y=5
77. 2x+6y=12 78. 5x+2y=10
79. x=3 80. x=-4

Everyday Math

81. Motor home cost. The Stonechilds rented a motor home for one week to go on vacation. It cost them $594 plus $0.32 per mile to rent the motor home, so the linear equation y=594+0.32x gives the cost, y, for driving x miles. Calculate the rental cost for driving 400, 800, and 1200 miles, and then graph the line. 82. Weekly earnings. At the art gallery where he works, Archisma gets paid $200 per week plus 15% of the sales he makes, so the equation y=200+0.15x gives the amount, y, he earns for selling x dollars of artwork. Calculate the amount Archisma earns for selling $900, $1600, and $2000, and then graph the line.

Writing Exercises

83. Explain how you would choose three x– values to make a table to graph the line y=1 over 5x-2. 84. What is the difference between the equations of a vertical and a horizontal line?

Answers

1. a) yes; no b) no; no c) yes; yes d) yes; yes 3. a) yes; yes b) yes; yes c) yes; yes d) no; no
5.

Graph of the equation y = 3x − 1.

 

7.

Graph of the equation y = −3x + 3.

9.

Graph of the equation y = x + 2.

11.

Graph of the equation y = −x − 3.

 

13.

Graph of the equation y = 2x.

15.

Graph of the equation y = 3x.

17.

Graph of the equation y = 1 half x + 2.

19.

Graph of the equation y = 4 thirds x − 5.

21.

Graph of the equation y = − 2 fifths x + 1.

23.

Graph of the equation y = − 3 halves x + 2.

25.

Graph of the equation x + y = 6.

27.

Graph of the equation x + y = −3.

29.

Graph of the equation x − y = 2.

31.

Graph of the equation x − y = −1.

33.

Graph of the equation 3x + y = 7.

35.

Graph of the equation 2x + y = −3.

 

37.

Graph of the equation 1 third x + y = 2.

39. *ANSWER GRAPH LOOKS OFF; ie. graph should have m=2/5, not (-2/5).
41.

Graph of the equation 2x + 3y = 12.

43.

Graph of the equation 3x − 4y = 12.

45.

Graph of the equation x − 6y = 3.

47.

Graph of the equation 3x + y = 2.

49.

Graph of the equation x = 4. The resulting line is vertical.

51.

Graph of the equation x = −2. The resulting line is vertical.

53.

Graph of the line y = 3. The resulting line is horizontal.

55.

Graph of the line y = −5. The resulting line is horizontal.

57.

Graph of the equation x = 7 thirds. The resulting line is vertical.

59.

Graph of the equation y = − 15 fourths. The resulting line is horizontal.

61.

The equations y= 2x and y = 2 are graphed. The equation y = 2x is a slanted line while y = 2 is horizontal.

63.

The equations y = − 1 half x and y = − 1 half are graphed. The equation y = − 1 half x is a slanted line while y = − 1 half is horizontal.

65.

Graph of the equation y = 4x.

67.

Graph of the equation y = − 1 half x + 3.

69.

Graph of the equation y = − x.

71.

graph of the equation x − y = 3.

73.

Graph of the equation 4x + y = 2.

75.

Graph of the equation y = −1.

77.

Graph of the equation 2x + 6y = 12.

79.

Graph of the equation x = 3.

81. $722, $850, $978
Graph of the equation y = 594 + 0.32x.
83. Answers will vary.

Attributions

This chapter has been adapted from “Graph Linear Equations in Two Variables” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

34

6.3 Graph with Intercepts

Learning Objectives

By the end of this section, you will be able to:

  • Identify the x– and y– intercepts on a graph
  • Find the x– and y– intercepts from an equation of a line
  • Graph a line using the intercepts

Identify the x– and y– Intercepts on a Graph

Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.

At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x– axis and the y– axis. These points are called the intercepts of the line.

Intercepts of a line

The points where a line crosses the x– axis and the y– axis are called the intercepts of a line.

Let’s look at the graphs of the lines in (Figure 1).

Examples of graphs crossing the x-negative axis.

Four figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line crossing the x- axis at the point (3, 0) and crossing the y- axis at the point (0, 6). The graph is labeled with the equation 2x plus y equals 6. Figure b shows a straight line crossing the x- axis at the point (4, 0) and crossing the y- axis at the point (0, negative 3). The graph is labeled with the equation 3x minus 4y equals 12. Figure c shows a straight line crossing the x- axis at the point (5, 0) and crossing the y- axis at the point (0, negative 5). The graph is labeled with the equation x minus y equals 5. Figure d shows a straight line crossing the x- axis and y- axis at the point (0, 0). The graph is labeled with the equation y equals negative 2x.
Figure .1

First, notice where each of these lines crosses the x negative axis. See (Figure 1).

Figure The line crosses the x– axis at: Ordered pair of this point
Figure (a) 3 (3,0)
Figure (b) 4 (4,0)
Figure (c) 5 (5,0)
Figure (d) 0 (0,0)

Do you see a pattern?

For each row, the y– coordinate of the point where the line crosses the x– axis is zero. The point where the line crosses the x– axis has the form (a,0) and is called the x– intercept of a line. The x– intercept occurs when y is zero.

Now, let’s look at the points where these lines cross the y– axis. See the table below.

Figure The line crosses the y-axis at: Ordered pair for this point
Figure (a) 6 (0,6)
Figure (b) -3 (0,-3)
Figure (c) -5 (0,5)
Figure (d) 0 (0,0)

What is the pattern here?

In each row, the x– coordinate of the point where the line crosses the y– axis is zero. The point where the line crosses the y– axis has the form (0,b) and is called the y- intercept of the line. The y– intercept occurs when x is zero.

x– intercept and y– intercept of a line

The x– intercept is the point (a,0) where the line crosses the x– axis.

The y– intercept is the point (0,b) where the line crosses the y– axis.

No Alt Text

EXAMPLE 1

Find the x– and y– intercepts on each graph.

Three figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line going through the points (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), and (6, negative 1). Figure b shows a straight line going through the points (0, negative 6), (1, negative 3), (2, 0), (3, 3), and (4, 6). Figure c shows a straight line going through the points (negative 6, 1), (negative 5, 0), (negative 4, negative 1), (negative 3, negative 2), (negative 2, negative 3), (negative 1, negative 4), (0, negative 5), and (1, negative 6).

Solution
  1. The graph crosses the x– axis at the point (4,0). The x– intercept is (4,0).
    The graph crosses the y– axis at the point (0,2). The y– intercept is (0,2).
  2. The graph crosses the x– axis at the point (2,0). The x– intercept is (2,0)
    The graph crosses the y– axis at the point (0,-6). The y– intercept is (0,-6).
  3. The graph crosses the x– axis at the point (-5,0). The x– intercept is (-5,0).
    The graph crosses the y– axis at the point (0,-5). The y– intercept is (0,-5).

TRY IT 1.1

Find the x– and y– intercepts on the graph.

Graph of the equation y = x − 2. The x-intercept is the point (2, 0) and the y-intercept is the point (0, −2)

Show answer

x– intercept: (2,0); y– intercept: (0,-2)

TRY IT 1.2

Find the x– and y– intercepts on the graph.

Graph of the equation y = − 2 thirds x + 2 and the x-intercept is the point (3, 0) and the y-intercept is the point (0, 2).

Show answer

x– intercept: (3,0), y– intercept: (0,2)

Find the x– and y– Intercepts from an Equation of a Line

Recognizing that the x– intercept occurs when y is zero and that the y– intercept occurs when x is zero, gives us a method to find the intercepts of a line from its equation. To find the x– intercept, let y=0 and solve for x. To find the y– intercept, let x=0 and solve for y.

Find the x– and y– intercepts from the equation of a line

Use the equation of the line. To find:

  • the x– intercept of the line, let y=0 and solve for x.
  • the y– intercept of the line, let x=0 and solve for y.

EXAMPLE 2

Find the intercepts of 2x+y=6.

Solution

We will let y=0 to find the x– intercept, and let x=0 to find the y– intercept. We will fill in the table, which reminds us of what we need to find.

The figure shows a table with four rows and two columns. The first row is a title row and it labels the table with the equation 2 x plus y equals 6. The second row is a header row and it labels each column. The first column header is “x” and the second is "y". The third row is labeled “x- intercept” and has the first column blank and a 0 in the second column. The fourth row is labeled “y- intercept” and has a 0 in the first column with the second column blank.

To find the x– intercept, let y=0.

.
Let y = 0. .
Simplify. .
.
The x-intercept is (3, 0)
To find the y-intercept, let x = 0.
.
Let x = 0. .
Simplify. .
.
The y-intercept is (0, 6)

The intercepts are the points (3,0) and (0,6) as shown in the following table.

2x+y=6
x y
3 0
0 6

TRY 2.1

Find the intercepts of 3x+y=12.

Show answer

x– intercept: (4,0), y– intercept: (0,12)

TRY IT 2.2

Find the intercepts of x+4y=8.

Show answer

x– intercept: (8,0), y– intercept: (0,2)

EXAMPLE 3

Find the intercepts of 4x-3y=12.

Solution
To find the x-intercept, let y = 0.
.
Let y = 0. .
Simplify. .
.
.
The x-intercept is (3, 0)
To find the y-intercept, let x = 0.
.
Let x = 0. .
Simplify. .
.
.
The y-intercept is (0, −4)

The intercepts are the points (3, 0) and (0, −4) as shown in the following table.

4x-3y=12
x y
3 0
0 -4

TRY IT 3.1

Find the intercepts of 3x-4y=12.

Show answer

x– intercept: (4,0), y– intercept: (0,-3)

TRY IT 3.2

Find the intercepts of 2x-4y=8.

Show answer

x– intercept: (4,0), y– intercept: (0,-2)

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x– and y– intercepts as two of your three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.

EXAMPLE 4

How to Graph a Line Using Intercepts

Graph -x+2y=6 using the intercepts.

Solution

The figure shows a table with the general procedure for graphing a line using the intercepts along with a specific example using the equation negative x plus 2y equals 6. Step 1 of the general procedure is “Find the x and y- intercepts of the line. Let y equals 0 and solve for x. Let x equals 0 and solve for y”. Step 1 for the example is a series of statements and equations: “Find the x- intercept. Let y equals 0”, negative x plus 2y equals 6, negative x plus 2(0) equals 6 (where the 0 is red), negative x equals 6, x equals negative 6, “The x- intercept is (negative 6, 0)”, “Find the y- intercept. Let x equals 0”, negative x plus 2y equals 6, negative 0 plus 2y equals 6 (where the 0 is red), 2y equals 6, y equals 3, and “The y- intercept is (0, 3)”.Step 2 of the general procedure is “Find another solution to the equation.” Step 2 for the example is a series of statements and equations: “We’ll use x equals 2”, “Let x equals 2”, negative x plus 2y equals 6, negative 2 plus 2y equals 6 (where the first 2 is red), 2y equals 8, y equals 4, and “A third point is (2, 4)”. Step 3 of the general procedure is “Plot the three points. Check that the points line up.”Step 3 for the example is a table and a graph. The table has four rows and three columns. The first row is a header row and it labels each column. The first column header is “x”, the second is "y", and the third is “(x,y)”. Under the first column are the numbers negative 6, 0 and 2. Under the second column are the numbers 0, 3, and 4. Under the third column are the ordered pairs (negative 6, 0), (0, 3), and (2, 4). The graph has three points on the x- y coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4).Step 4 of the general procedure is “Draw the line.” For the specific example, there is the statement “See the graph” and a graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4). The straight line is drawn through the points (negative 6, 0), (negative 4, 1), (negative 2, 2), (0, 3), (2, 4), (4, 5), and (6, 6).

TRY IT 4.1

Graph x-2y=4 using the intercepts.

Show answer
Graph of the equation x − 2y = 4. The x-intercept is the point (4, 0) and the y-intercept is the point (0, −2).

TRY IT 4.2

Graph -x+3y=6 using the intercepts.

Show answer
Graph of the equation −x + 3y = 6. The x-intercept is the point (−6, 0) and the y-intercept is the point (0, 2).

HOW TO: Graph a linear equation using the intercepts

The steps to graph a linear equation using the intercepts are summarized below.

  1. Find the x– and y– intercepts of the line.
    • Let y=0 and solve for x
    • Let x=0 and solve for y.
  2. Find a third solution to the equation.
  3. Plot the three points and check that they line up.
  4. Draw the line.

EXAMPLE 5

Graph 4x-3y=12 using the intercepts.

Solution

Find the intercepts and a third point.

The figure shows a series of statements and equations: “Find the x- intercept. Let y equals 0”, 4x minus 3y equals 12, 4x minus 3(0) equals 12 (where the 0 is red), 4x equals 12, x equals 3, “Find the y- intercept. Let x equals 0”, 4x minus 3y equals 12, 4(0) minus 3y equals 12 (where the 0 is red), negative 3y equals 12, y equals negative 4, “third point, let y equals 4”, 4x minus 3y equals 12, 4x minus 3(4) equals 12 (where the second 4 is red), 4x minus 12 equals 12, 4x equals 24, and x equals 6.

We list the points in following table and show the graph below.

4x-3y=12
x y (x,y)
3 0 (3,0)
0 -4 (0,-4)
6 4 (6,4)

The points listed on the previous table are plotted. The equation graphed is 4x − 3y = 12.

TRY IT 5.1

Graph 5x-2y=10 using the intercepts.

Show answer
Graph of the equation 5x − 2y = 10.

TRY IT 5.2

Graph 3x-4y=12 using the intercepts.

Show answer
Graph of the equation 3x − 4y = 12.

EXAMPLE 6

Graph y=5x using the intercepts.

Solution

The figure shows two sets of statements and equations to find the intercepts from an equation. The first set of statements and equations is “x- intercept”, “let y equals 0”, y equals 5x, 0 equals 5x (where the 0 is red), 0 equals x, (0, 0). The second set of statements and equations is “y- intercept”, “let x equals 0”, y equals 5x, y equals 5(0) (where the 0 is red), y equals 0, (0, 0).

This line has only one intercept. It is the point (0,0).

To ensure accuracy we need to plot three points. Since the x– and y– intercepts are the same point, we need two more points to graph the line.

The figure shows two sets of statements and equations to find two points from an equation. The first set of statements and equations is “Let x equals 1”, y equals 5x, y equals 5(1) (where the 1 is red), y equals 5. The second set of statements and equations is “Let x equals negative 1”, y equals 5x, y equals 5(negative 1) (where the negative 1 is red), y equals negative 5.

See following table..

y=5x
x y (x,y)
0 0 (0,0)
1 5 (1,5)
-1 -5 (-1,-5)

Plot the three points, check that they line up, and draw the line.

The points from the previous table are plotted and labeled. The equation graphed is y = 5x.

 

TRY IT 6.1

Graph y=4x using the intercepts.

Show answer
Graph of the equation y = 4x.

TRY IT 6.2

Graph y=-x the intercepts.

Show answer
Graph of the equation y = −x.

 

Key Concepts

  • Find the x– and y– Intercepts from the Equation of a Line
    • Use the equation of the line to find the x– intercept of the line, let y=0 and solve for x.
    • Use the equation of the line to find the y– intercept of the line, let x=0 and solve for y.
  • Graph a Linear Equation using the Intercepts
    1. Find the x– and y– intercepts of the line.
      Let y=0 and solve for x.
      Let x=0 and solve for y.
    2. Find a third solution to the equation.
    3. Plot the three points and then check that they line up.
    4. Draw the line.
  • Strategy for Choosing the Most Convenient Method to Graph a Line:
    • Consider the form of the equation.
    • If it only has one variable, it is a vertical or horizontal line.
      x=a is a vertical line passing through the x– axis at a
      y=b is a horizontal line passing through the y– axis at b.
    • If y is isolated on one side of the equation, graph by plotting points.
    • Choose any three values for x and then solve for the corresponding y– values.
    • If the equation is of the form ax+by=c, find the intercepts. Find the x– and y– intercepts and then a third point.

Glossary

intercepts of a line
The points where a line crosses the x– axis and the y– axis are called the intercepts of the line.
x– intercept
The point (a,0) where the line crosses the x– axis; the x– intercept occurs when y is zero.
y-intercept
The point (0,b) where the line crosses the y– axis; the y– intercept occurs when x is zero.

Practice Makes Perfect

Identify the x– and y– Intercepts on a Graph

In the following exercises, find the x– and y– intercepts on each graph.

1.
Graph of the equation y = −x +3. The x-intercept is the point (3, 0) and the y-intercept is the point (0, 3).
2.
The graph of the equation y = −x + 2. The x-intercept is the point (2, 0) and the y-intercept is the point (0, 2).
3.
Graph of the equation y = x − 5. The x-intercept is the point (5, 0) and the y-intercept is the point (0, −5).

4.

Graph of the equation y = x − 1. The x-intercept is the point (1, 0) and the y-intercept is the point (0, −1)
5.
Graph of the equation y = −x − 2. The x-intercept is the point (−2, 0) and the y-intercept is the point (−2, 0).

6.

Graph of the equation y = −x − 3. The x-intercept is the point (−3, 0) and the y-intercept is the point (0, −3).
7.
Graph of the equation y = x + 1. The x-intercept is the point (−1, 0) and the y-intercept is the point (0, 1).

8.

Graph of the equation y = x + 5. The x-intercept is point (−5, 0) and the y-intercept is the point (0, 5).
9.
Graph of the equation y = − 1 half x + 3. The x-intercept is the point (6, 0) and the y-intercept is the point (0, 3).

10.

Graph of the equation y = − 1 half x + 2. The x-intercept is the point (4, 0) and the y-intercept is the point (0, 2).
11.
Graph of the equation y = x. Both the x-intercept and y-intercept is the point (0, 0).

12.

Graph of the equation y = x. Both the x-intercept and y-intercept is the point (0, 0).

Find the x– and y– Intercepts from an Equation of a Line

In the following exercises, find the intercepts for each equation.

13. x+y=4 14. x+y=3
15. x+y=-2 17. x-y=5
18. x-y=1 19. x-y=-3
20. x-y=-4 21. x+2y=8
22. x+2y=10 23. 3x+y=6
24. 3x+y=9 25. x-3y=12
25. x-3y=12 27. 4x-y=8
28. 5x-y=5 28. 5x-y=5
30. 2x+3y=6 31. 3x-2y=12
32. 3x-5y=30 33. y=1 over 3x+1
34. y=1 over 4x-1 35. y=1 over 5x+2
36. y=1 over 3x+4 37. y=3x
38. y=-2x 39. y=-4x
40. y=5x

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

41. -x+5y=10 42. -x+4y=8
43. x+2y=4 44. x+2y=6
45. x+y=2 46. x+y=5
47. x+y=-3 48. x+y=-1
49. x-y=1 49. x-y=1
51. x-y=-4 52. x-y=-3
53. 4x+y=4 54. 3x+y=3
55. 2x+4y=12 56. 3x+2y=12
57. 3x-2y=6 58. 5x-2y=10
59. 2x-5y=-20 60. 3x-4y=-12
61. 3x-y=-6 62. 2x-y=-8
63. y=3 over 2x 64. y=-4x
65. y=x 66. y=3x

Everyday Math

67. Road trip. Damien is driving from Thunder Bay to Montreal, a distance of 1000 miles. The x– axis on the graph below shows the time in hours since Damien left Thunder Bay. The y– axis represents the distance he has left to drive.

Points plotted and labeled on the graph are described in the previous paragraph. A line is drawn between the points.

  1. a) Find the x– and y– intercepts.
  2. b) Explain what the x– and y– intercepts mean for Damien.

68. Road trip. Jenna filled up the gas tank of her truck and headed out on a road trip. The x– axis on the graph below shows the number of miles Jenna drove since filling up. The y– axis represents the number of gallons of gas in the truck’s gas tank.

Points plotted and labeled on the graph are described in the previous paragraph. A line is drawn between the points.

  1. a) Find the x– and y– intercepts.
  2. b) Explain what the x– and y– intercepts mean for Ozzie.

Writing Exercises

69. How do you find the x– intercept of the graph of 3x-2y=6? 70. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation 4x+y=-4? Why?
71. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation y=2 over 3x-2? Why? 72. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation y=6? Why?

Answers

1. (3,0),(0,3) 3. (5,0),(0,-5)
5. (-2,0),(0,-2) 7. (-1,0),(0,1)
9. (6,0),(0,3) 11. (0,0)
13. (4,0),(0,4) 15. (-2,0),(0,-2)
17. (5,0),(0,-5) 19. mathematical expression
21. (8,0),(0,4) 23. (2,0),(0,6)
25. (12,0),(0,-4) 27. (2,0),(0,-8)
29. (5,0),(0,2) 31. (4,0),(0,-6)
33. (-3,0),(0,1) 35. (-10,0),(0,2)
37. (0,0) 39. (0,0)
41.

Graph of the equation −x + 5y = 10. The x-intercept is the point (−10, 0) and the y-intercept is the point (0, 2).

43.

Graph of the equation x + 2 = 4. The x-intercept is the point (4, 0) and the y-intercept is the point (0, 2).

45.

Graph of the equation x + y = 2. The x-intercept is the point (2, 0) and the y-intercept is the point (0, 2).

47.

Graph of the equation x + y = −3. The x-intercept is the point (−3, 0) and the y-intercept is the point (0, −3).

49.

Graph of the equation x − y = 1. The x-intercept is the point (1, 0) is the y-intercept is the point (0, −1).

51.

Graph of the equation x − y = −4. The x-intercept is the point (−4, 0) and the y-intercept is the point (0, 4).

53.

Graph of the equation 4x + y = 4. The x-intercept is the point (1, 0) and the y-intercept is the point (0, 4).

 

55.

Graph of the equation 2x + 4y = 12. The x-intercept is the point (6, 0) and the y-intercept is the point (0, 3).

57.

Graph of the equation 3x − 2y = 6. The x-intercept is the point (2, 0) and the y-intercept is the point (−3, 0).

59.

Graph of the equation 2x − 5y = −20. The x-intercept is the point (−10, 0) and the y-intercept is the point (4, 0).

61.

Graph of the equation 3x − y = −6. The x-intercept is the point (−2, 0) and the y-intercept is the point (0, 6).

63.

Graph of the equation y = 3 halves x − 3. The x-intercept is the point (2, 0) and the y-intercept is the point (0, −3).

65.

Graph of the equation y = x. Both the x-intercept and the y-intercept is the point (0, 0).

67.

a)(0,1000),(15,0)
b) At (0,1000), he has been gone 0 hours and has 1000 miles left. At (15,0), he has been gone 15 hours and has 0 miles left to go.

69. Answers will vary. 71. Answers will vary.

Attributions

This chapter has been adapted from “Graph with Intercepts” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

35

6.4 Understand Slope of a Line

Learning Objectives

By the end of this section, you will be able to:

  • Use geoboards to model slope
  • Use m=rise over run to find the slope of a line from its graph
  • Find the slope of horizontal and vertical lines
  • Use the slope formula to find the slope of a line between two points
  • Graph a line given a point and the slope
  • Solve slope applications

When you graph linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter. What determines whether a line tilts up or down or if it is steep or flat?

In mathematics, the ‘tilt’ of a line is called the slope of the line. The concept of slope has many applications in the real world. The pitch of a roof, grade of a highway, and a ramp for a wheelchair are some examples where you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.

In this section, we will explore the concept of slope.

Use Geoboards to Model Slope

A geoboard is a board with a grid of pegs on it. Using rubber bands on a geoboard gives us a concrete way to model lines on a coordinate grid. By stretching a rubber band between two pegs on a geoboard, we can discover how to find the slope of a line.

Doing the Manipulative Mathematics activity “Exploring Slope” will help you develop a better understanding of the slope of a line. (Graph paper can be used instead of a geoboard, if needed.)

We’ll start by stretching a rubber band between two pegs as shown in (Figure 1).

A 5 by 5 grid of pegs. A rubbed band is stretched between two pegs, forming a line.
Figure .1

Doesn’t it look like a line?

Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle, as shown in (Figure 2)

A 5 by 5 grid of pegs. A rubbed band is stretched between three pegs, forming 3 lines that are connected to each other.
Figure .2

We carefully make a 90º angle around the third peg, so one of the newly formed lines is vertical and the other is horizontal.

To find the slope of the line, we measure the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in (Figure 3).

A vertical arrow that is labeled “rise” and a horizontal arrow that is labeled “run”.
Figure .3

If our geoboard and rubber band look just like the one shown in (Figure 4), the rise is 2. The rubber band goes up 2 units. (Each space is one unit.)

The rise on this geoboard is 2, as the rubber band goes up two units.

The same picture as Figure .3 except the vertical “rise” line is labeled 2 and the horizontal “run” line is labeled 3.
Figure .4

What is the run?

The rubber band goes across 3 units. The run is 3 (see (Figure 4)).

The slope of a line is the ratio of the rise to the run. In mathematics, it is always referred to with the letter m.

Slope of a line

The slope of a line of a line is m=rise over run.

The rise measures the vertical change and the run measures the horizontal change between two points on the line.

What is the slope of the line on the geoboard in (Figure 4)?

mathematical expression

The line has slope 2 over 3. This means that the line rises 2 units for every 3 units of run.

When we work with geoboards, it is a good idea to get in the habit of starting at a peg on the left and connecting to a peg to the right. If the rise goes up it is positive and if it goes down it is negative. The run will go from left to right and be positive.

EXAMPLE 1

What is the slope of the line on the geoboard shown?

A 5 by 5 grid of pegs. A rubber band stretched between the pegs (1, 5) and (5, 2).

Solution

Use the definition of slope: m=rise over run.

Start at the left peg and count the spaces up and to the right to reach the second peg.

5 by 5 grid of pegs. A rubber band stretched between pegs (1, 1), (5, 2), and (1, 4). Horizontal is "4", vertical is "3".

The rise is 3. m=3 over run
The run is 4. m=3 over 4
The slope is 3 over 4.

This means that the line rises 3 units for every 4 units of run.

TRY IT 1.1

What is the slope of the line on the geoboard shown?

A 5 by 5 grid of pegs. A rubber band is stretched between the pegs (1,1) and (5, 4).

Show answer

4 over 3

TRY IT 1.2

What is the slope of the line on the geoboard shown?

A 5 by 5 grid of pegs. A rubber band is stretched between the pegs (1, 2) and (5, 3).

Show answer

1 over 4

EXAMPLE 2

What is the slope of the line on the geoboard shown?

A 5 by 5 grid of pegs. A rubber band stretched between the pegs (1, 3) and (4, 2).

Solution

Use the definition of slope: m=rise over run.

Start at the left peg and count the units down and to the right to reach the second peg.

5 by 5 grid of pegs. A rubber band stretched between pegs (1, 3), (4, 2), and (1, 2). Horizontal is “3”, vertical is “−1".

The rise is −1. =-1 over run
The run is 3. mathematical expression
The slope is -1 over 3.

This means that the line drops 1 unit for every 3 units of run.

TRY IT 2.1

What is the slope of the line on the geoboard?

Show answer

-2 over 3

TRY IT 2.2

What is the slope of the line on the geoboard?

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 1 and the peg in column 4, row 5, forming a line.

Show answer

-4 over 3

Notice that in (Example 1) the slope is positive and in (Example 2) the slope is negative. Do you notice any difference in the two lines shown in (Figure 5a) and (Figure 5b)?

The figure shows two grids of evenly spaced pegs, one labeled (a) and one labeled (b). There are 5 columns and 5 rows of pegs in each grid. In the (a) grid, a rubber band is stretched between the peg in column 1, row 5 and the peg in column 5, row 2, forming a line. Below this grid is the slope of a line defined as m equals 3 fourths. In the (b) grid, a rubber band is stretched between the peg in column 1, row 3 and the peg in column 4, row 4, forming a line. Below this grid is the slope of a line defined as m equals negative 1 third.
Figure .5 (a) (b)

Positive and negative slopes

We ‘read’ a line from left to right just like we read words in English. As you read from left to right, the line in (Figure 5a) is going up; it has positive slope. The line in (Figure 5b) is going down; it has negative slope.

The figure shows two lines side-by-side. The line on the left is a diagonal line that rises from left to right. It is labeled “Positive slope”. The line on the right is a diagonal line that drops from left to right. It is labeled “Negative slope”.

EXAMPLE 3

Use a geoboard to model a line with slope 1 over 2.

Solution

To model a line on a geoboard, we need the rise and the run.

Use the slope formula. m=rise over run
Replace m with 1 over 2. 1 over 2=rise over run

So, the rise is 1 and the run is 2

Start at a peg in the lower left of the geoboard.

Stretch the rubber band up 1 unit, and then right 2 units.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 3, the peg in column 1, row 4 and the peg in column 3, row 3, forming a right triangle. The 1, 3 peg forms the vertex of the 90 degree angle and the line from the 1, 4 peg to the 3, 3 peg forms the hypotenuse of the triangle. The line from the 1, 3 peg to the 1, 4 peg is labeled “1”. The line from the 1, 3 peg to the 3, 3 peg is labeled “2”.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is 1 over 2.

TRY IT 3.1

Model the slope m=1 over 3. Draw a picture to show your results.

Show answer
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 3, the peg in column 2, row 4 and the peg in column 5, row 3, forming a right triangle. The 2, 3 peg forms the vertex of the 90 degree angle and the line from the 2, 4 peg to the 5, 3 peg forms the hypotenuse of the triangle.

TRY IT 3.2

Model the slope m=3 over 2. Draw a picture to show your results.

Show answer
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 1, the peg in column 1, row 4 and the peg in column 3, row 1, forming a right triangle. The 1, 1 peg forms the vertex of the 90 degree angle and the line from the 1, 4 peg to the 3, 1 peg forms the hypotenuse of the triangle.

EXAMPLE 4

Use a geoboard to model a line with slope -1 over 4.

Solution
Use the slope formula. m=rise over run
Replace m with -1 over 4. -1 over 4=rise over run

So, the rise is -1 and the run is 4

Since the rise is negative, we choose a starting peg on the upper left that will give us room to count down.

We stretch the rubber band down 1 unit, then go to the right 4 units, as shown.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 2, the peg in column 1, row 3 and the peg in column 5, row 3, forming a right triangle. The 1, 3 peg forms the vertex of the 90 degree angle and the line from the 1, 2 peg to the 5, 3 peg forms the hypotenuse of the triangle. The line from the 1, 2 peg to the 1, 3 peg is labeled “negative 1”. The line from the 1, 3 peg to the 5, 3 peg is labeled “4”.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is -1 over 4.

TRY IT 4.1

Model the slope m=-2 over 3. Draw a picture to show your results.

Show answer
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 3, the peg in column 2, row 5 and the peg in column 3, row 5, forming a right triangle. The 2, 5 peg forms the vertex of the 90 degree angle and the line from the 2, 3 peg to the 3, 5 peg forms the hypotenuse of the triangle.

TRY IT 4.2

Model the slope m=-1 over 3. Draw a picture to show your results.

Show answer
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 1, the peg in column 1, row 2 and the peg in column 4, row 2, forming a right triangle. The 1, 2 peg forms the vertex of the 90 degree angle and the line from the 1, 1 peg to the 4, 2 peg forms the hypotenuse of the triangle.

Use m=rise over run to Find the Slope of a Line from its Graph

Now, we’ll look at some graphs on the xy-coordinate plane and see how to find their slopes. The method will be very similar to what we just modeled on our geoboards.

To find the slope, we must count out the rise and the run. But where do we start?

We locate two points on the line whose coordinates are integers. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.

EXAMPLE 5

How to Use m=rise over run to Find the Slope of a Line from its Graph

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 6 and the y-axis runs from negative 4 to 2. A line passes through the points (0, negative 3) and (5, 1).

Solution

This table has three columns and four rows. The first row says, “Step 1. Locate two points on the graph whose coordinates are integers. Mark (0, negative 3) and (5, 1).” To the right is a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 6. The y-axis of the plane runs from negative 4 to 2. The points (0, negative 3) and (5, 1) are plotted.The second row says, “Step 2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point. Starting at (0, negative 3), sketch a right triangle to (5, 1).” In the graph on the right, an additional point is plotted at (0, 1). The three points form a right triangle, with the line from (0, negative 3) to (5, 1) forming the hypotenuse and the lines from (0, negative 3) to (0, 1) and (0, 1) to (5, 1) forming the legs.The third row then says, “Step 3. Count the rise and the run on the legs of the triangle.” The rise is 4 and the run is 5.The fourth row says, “Step 4. Take the ratio of the rise to run to find the slope. Use the slope formula. Substitute the values of the rise and run.” To the right is the slope formula, m equals rise divided by run. The slope of the line is 4 divided by 5, or four fifths. This means that y increases 4 units as x increases 5 units.

TRY IT 5.1

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 8 to 1 and the y-axis runs from negative 1 to 4. A line passes through the points (negative 5, 1) and (0, 3).

Show answer

2 over 5

TRY IT 5.2

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 2 to 4. A line passes through the points (0, negative 1) and (4, 2).

Show answer

3 over 4

HOW TO: Find the slope of a line from its graph using mathematical expression.

  1. Locate two points on the line whose coordinates are integers.
  2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
  3. Count the rise and the run on the legs of the triangle.
  4. Take the ratio of rise to run to find the slope, m=rise over run.

EXAMPLE 6

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 9 and the y-axis runs from negative 1 to 7. A line passes through the points (0, 5), (3, 3), and (6, 1).

Solution
Locate two points on the graph whose coordinates are integers. (0,5) and (3,3)
Which point is on the left? (0,5)
Starting at (0,5), sketch a right triangle to (3,3). .
Count the rise—it is negative. The rise is -2.
Count the run. The run is 3.
Use the slope formula. m=rise over run
Substitute the values of the rise and run. m=-2 over 3
Simplify. m=-2 over 3
The slope of the line is -2 over 3.

So y increases by 3 units as x decreases by 2 units.

What if we used the points (-3,7) and (6,1) to find the slope of the line?

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 3, 7) and (6, 1). An additional point is plotted at (negative 3, 1). The three points form a right triangle, with the line from (negative 3, 7) to (6, 1) forming the hypotenuse and the lines from (negative 3, 7) to negative 1, 7) and from (negative 1, 7) to (6, 1) forming the legs.

The rise would be -6 and the run would be 9. Then m=-6 over 9, and that simplifies to m=-2 over 3. Remember, it does not matter which points you use—the slope of the line is always the same.

TRY IT 6.1

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 6 to 1. A line passes through the points (0, negative 2) and (3, negative 6).

Show answer

-4 over 3

TRY IT 6.2

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 3 to 6 and the y-axis runs from negative 3 to 2. A line passes through the points (0, 1) and (5, negative 2).

Show answer

-3 over 5

In the last two examples, the lines had y-intercepts with integer values, so it was convenient to use the y-intercept as one of the points to find the slope. In the next example, the y-intercept is a fraction. Instead of using that point, we’ll look for two other points whose coordinates are integers. This will make the slope calculations easier.

EXAMPLE 7

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from 0 to 8 and the y-axis runs from 0 to 7. A line passes through the points (2, 3) and (7, 6).

Solution
Locate two points on the graph whose coordinates are integers. (2,3) and (7,6)
Which point is on the left? (2,3)
Starting at (2,3), sketch a right triangle to (7,6). .
Count the rise. The rise is 3.
Count the run. The run is 5.
Use the slope formula. m=rise over run
Substitute the values of the rise and run. m=3 over 5
The slope of the line is 3 over 5.

This means that y increases 5 units as x increases 3 units.

When we used geoboards to introduce the concept of slope, we said that we would always start with the point on the left and count the rise and the run to get to the point on the right. That way the run was always positive and the rise determined whether the slope was positive or negative.

What would happen if we started with the point on the right?

Let’s use the points (2,3) and (7,6) again, but now we’ll start at (7,6).

The graph shows the x y coordinate plane. The x -axis runs from 0 to 8. The y -axis runs from 0 to 7. A line passes through the points (2, 3) and (7, 6). An additional point is plotted at (7, 3). The three points form a right triangle, with the line from (2, 3) to (7, 6) forming the hypotenuse and the lines from (2, 3) to (7, 3) and from (7, 3) to (7, 6) forming the legs.

Count the rise. The rise is -3.
Count the run. It goes from right to left, so it is negative. The run is -5.
Use the slope formula. m=rise over run
Substitute the values of the rise and run. m=-3 over -5
The slope of the line is -3 over -5.

It does not matter where you start—the slope of the line is always the same.

TRY IT 7.1

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 4 to 2 and the y-axis runs from negative 6 to 2. A line passes through the points (negative 3, 4) and (1, 1).

Show answer

5 over 4

EXAMPLE 7.2

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 4 and the y-axis runs from negative 2 to 3. A line passes through the points (1, negative 1) and (3, 2).

Show answer

3 over 2

Find the Slope of Horizontal and Vertical Lines

Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.

mathematical expression

So how do we find the slope of the horizontal line y=4? One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens when we do this.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 1 to 7. A line passes through the points (0, 4) and (3, 4).

What is the rise? The rise is 0.
Count the run. The run is 3.
What is the slope? mathematical expression
The slope of the horizontal line y=4 is 0.

All horizontal lines have slope 0. When the y-coordinates are the same, the rise is 0.

Slope of a horizontal line

The slope of a horizontal line, y=b, is 0.

The floor of your room is horizontal. Its slope is 0. If you carefully placed a ball on the floor, it would not roll away.

Now, we’ll consider a vertical line, the line.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 2 to 2. A line passes through the points (3, 0) and (3, 2).

What is the rise? The rise is 2.
Count the run. The run is 0.
What is the slope? mathematical expression

But we can’t divide by 0. Division by 0 is not defined. So we say that the slope of the vertical line x=3 is undefined.

The slope of any vertical line is undefined. When the x-coordinates of a line are all the same, the run is 0.

Slope of a vertical line

The slope of a vertical line, x=a, is undefined.

EXAMPLE 8

Find the slope of each line:

a) x=8 b) y=-5.

Solution

a) x=8
This is a vertical line.
Its slope is undefined.

b) y=-5
This is a horizontal line.
It has slope 0.

TRY IT 8.1

Find the slope of the line: x=-4.

Show answer

undefined

TRY 8.2

Find the slope of the line: y=7.

Show answer

0

Quick guide to the slopes of lines

This figure shows four lines with arrows. The first line rises up and runs to the right. It has a positive slope. The second line falls down and runs to the right. It has a negative slope. The third line is neither rises nor falls, extending horizontally in either direction. It has a slope of zero. The fourth line is completely vertical, one end rising up and the other rising down, running neither to the left nor right. It has an undefined slope.

Remember, we ‘read’ a line from left to right, just like we read written words in English.

Use the Slope Formula to find the Slope of a Line Between Two Points

Sometimes we’ll need to find the slope of a line between two points when we don’t have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but as we’ll see, there is a way to find the slope without graphing. Before we get to it, we need to introduce some algebraic notation.

We have seen that an ordered pair (x,y) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol (x,y) be used to represent two different points? Mathematicians use subscripts to distinguish the points.

mathematical expression

The use of subscripts in math is very much like the use of last name initials in elementary school. Maybe you remember Laura C. and Laura M. in your third grade class?

We will use (x sub 1,y sub 1) to identify the first point and (x sub 2,y sub 2) to identify the second point.

If we had more than two points, we could use (x sub 3,y sub 3), (x sub 4,y sub 4), and so on.

Let’s see how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points (2,3) and (7,6).

The graph shows the x y coordinate plane. The x and y-axes run from 0 to 7. A line passes through the points (2, 3) and (7, 6), which are plotted and labeled. The ordered pair (2, 3) is labeled (x subscript 1, y subscript 1). The ordered pair (7, 6) is labeled (x subscript 2, y subscript 2). An additional point is plotted at (2, 6). The three points form a right triangle, with the line from (2, 3) to (7, 6) forming the hypotenuse and the lines from (2, 3) to (2, 6) and from (2, 6) to (7, 6) forming the legs. The first leg, from (2, 3) to (2, 6) is labeled y subscript 2 minus y subscript 1, 6 minus 3, and 3. The second leg, from (2, 3) to (7, 6), is labeled x subscript 2 minus x subscript 1, y minus 2, and 5.

Since we have two points, we will use subscript notation, mathematical expressionmathematical expression.

On the graph, we counted the rise of 3 and the run of 5

Notice that the rise of 3 can be found by subtracting the y-coordinates 6 and 3

3=6-3

And the run of 5 can be found by subtracting the x-coordinates 7 and 2

5=7-2

We know m=rise over run. So m=3 over 5.

We rewrite the rise and run by putting in the coordinates m=6-3 over 7-2.

But 6 is y sub 2, the y-coordinate of the second point and 3 is y sub 1, the y-coordinate of the first point.

So we can rewrite the slope using subscript notation. m=y sub 2-y sub 1 over 7-2

Also, 7 is x sub 2, the x-coordinate of the second point and 2 is x sub 1, the x-coordinate of the first point.

So, again, we rewrite the slope using subscript notation. m=y sub 2-y sub 1 over x sub 2-x sub 1

We’ve shown that m=y sub 2-y sub 1 over x sub 2-x sub 1 is really another version of m=rise over run. We can use this formula to find the slope of a line when we have two points on the line.

Slope formula

The slope of the line between two points (x sub 1,y sub 1) and (x sub 2,y sub 2) is

m=y sub 2-y sub 1 over x sub 2-x sub 1

This is the slope formula.

The slope is:

mathematical expression

EXAMPLE 9

Use the slope formula to find the slope of the line between the points (1,2) and (4,5).

Solution
We’ll call (1,2) point #1 and (4,5) point #2. mathematical expressionmathematical expression.
Use the slope formula. m=y sub 2-y sub 1 over x sub 2-x sub 1.
Substitute the values.
y of the second point minus y of the first point m=5-2 over x sub 2-x sub 1.
x of the second point minus x of the first point m=5-2 over 4-1.
Simplify the numerator and the denominator. m=3 over 3.
Simplify. m=1.

Let’s confirm this by counting out the slope on a graph using m=rise over run.

The graph shows the x y-coordinate plane. The x and y-axes of the plane run from 0 to 7. A line passes through the points (1, 2) and (4, 5), which are plotted. An additional point is plotted at (1, 5). The three points form a right triangle, with the line from (1, 2) to (4, 5) forming the hypotenuse and the lines from (1, 2) to (1, 5) and from (1, 5) to (4, 5) forming the legs. The leg from (1, 2) to (1, 5) is labeled “rise” and the leg from (1, 5) to (4, 5) is labeled “run”.

It doesn’t matter which point you call point #1 and which one you call point #2. The slope will be the same. Try the calculation yourself.

TRY IT 9.1

Use the slope formula to find the slope of the line through the points: (8,5) and (6,3).

Show answer

1

TRY IT 9.2

Use the slope formula to find the slope of the line through the points: (1,5) and (5,9).

Show answer

1

EXAMPLE 10

Use the slope formula to find the slope of the line through the points (-2,-3) and (-7,4).

Solution
We’ll call (-2,-3) point #1 and (-7,4) point #2. mathematical expressionmathematical expression.
Use the slope formula. m=y sub 2-y sub 1 over x sub 2-x sub 1.
Substitute the values.
y of the second point minus y of the first point m=4-(-3) over x sub 2-x sub 1.
x of the second point minus x of the first point m=4-(-3) over -7-(-2).
Simplify. mathematical expression

Let’s verify this slope on the graph shown.

The graph shows the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 2 and the y-axis of the plane runs from negative 6 to 5. A line passes through the points (negative 7, 4) and (negative 2, negative 3), which are plotted and labeled. An additional point is plotted at (negative 7, negative 3). The three points form a right triangle, with the line from (negative 7, 4) to (negative 2, negative 3) forming the hypotenuse and the lines from (negative 7, 4) to (negative 7, negative 3) and from (negative 7, negative 3) to (negative 2, negative 3) forming the legs. The leg from (negative 7, 4) to (negative 7, negative 3) is labeled “rise” and the leg from (negative 7, negative 3) to (negative 2, negative 3) is labeled “run”.

mathematical expression

TRY IT 10.1

Use the slope formula to find the slope of the line through the points: (-3,4) and (2,-1).

Show answer

-1

TRY IT 10.2

Use the slope formula to find the slope of the line through the pair of points: (-2,6) and (-3,-4).

Show answer

10

Graph a Line Given a Point and the Slope

Up to now, in this chapter, we have graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

One other method we can use to graph lines is called the point–slope method. We will use this method when we know one point and the slope of the line. We will start by plotting the point and then use the definition of slope to draw the graph of the line.

EXAMPLE 11

How To Graph a Line Given a Point and The Slope

Graph the line passing through the point (1,-1) whose slope is m=3 over 4.

Solution

This table has three columns and four rows. The first row says, “Step 1. Plot the given point. Plot (1, negative 1).” To the right is a graph of the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 7. The y-axis of the plane runs from negative 3 to 4. The point (0, negative 1) is plotted.The second row says, “Step 2. Use the slope formula m equals rise divided by run to identify the rise and the run.” The rise and run are 3 and 4, so m equals 3 divided by 4.The third row says “Step 3. Starting at the given point, count out the rise and run to mark the second point.” We start at (1, negative 1) and count the rise and run. Up three units and right 4 units. In the graph on the right, an additional two points are plotted: (1, 2), which is 3 units up from (1, negative 1), and (5, 2), which is 3 units up and 4 units right from (1, negative 1).The fourth row says “Step 4. Connect the points with a line.” On the graph to the right, a line is drawn through the points (1, negative 1) and (5, 2). This line is also the hypotenuse of the right triangle formed by the three points, (1, negative 1), (1, 2) and (5, 2).

EXAMPLE 11.1

Graph the line passing through the point (2,-2) with the slope m=4 over 3.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 4, negative 10) and (2, negative 2).

TRY IT 11.2

Graph the line passing through the point (-2,3) with the slope m=1 over 4.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 2, 3) and (10, 6).

Graph a line given a point and the slope.

  1. Plot the given point.
  2. Use the slope formula m=rise over run to identify the rise and the run.
  3. Starting at the given point, count out the rise and run to mark the second point.
  4. Connect the points with a line.

EXAMPLE 12

Graph the line with y-intercept 2 whose slope is m=-2 over 3.

Solution

Plot the given point, the y-intercept, (0,2).

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The point (0, 2) is plotted.

Identify the rise and the run. m=-2 over 3
rise over run=-2 over 3
rise=-2
run=3

Count the rise and the run. Mark the second point.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The points (0, 2), (0, 0), and (3,0) are plotted and labeled. The line from (0, 2) to (0, 0) is labeled “down 2” and the line from (0, 0) to (3, 0) is labeled “right 3”.

Connect the two points with a line.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. A line passes through the plotted points (0, 2) and (3,0).

You can check your work by finding a third point. Since the slope is m=-2 over 3, it can be written as m=2 over -3. Go back to (0,2) and count out the rise, 2, and the run, -3.

TRY IT 12.1

Graph the line with the y-intercept 4 and slope m=-5 over 2.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the y-axis at (0, 4) and passes through the point (4, negative 6).

TRY IT 12.2

Graph the line with the x-intercept -3 and slope m=-3 over 4.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the x-axis at (negative 3, 0) and passes through the point (1, negative 3).

EXAMPLE 13

Graph the line passing through the point (-1,-3) whose slope is m=4.

Solution

Plot the given point.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The point (negative 1, negative 3) is plotted and labeled.

Identify the rise and the run. m=4
Write 4 as a fraction. rise over run=4 over 1
rise=4,run=1

Count the rise and run and mark the second point.

This figure shows how to graph the line passing through the point (negative 1, negative 3) whose slope is 4. The first step is to identify the rise and run. The rise is 4 and the run is 1. 4 divided by 1 is 4, so the slope is 4. Next we count the rise and run and mark the second point. To the right is a graph of the x y-coordinate plane. The x and y-axes run from negative 5 to 5. We start at the plotted point (negative 1, negative 3) and count the rise, 4. We reach the point negative 1, 1, which we plot. We then count the run from this point, which is 1. We reach the point (0, 1), which is plotted. The last step is to connect the two points with a line. We draw a line which passes through the points (negative 1, negative 3) and (0, 1).

Connect the two points with a line.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. A line passes through the plotted points (-1, -3) and (1,0).

You can check your work by finding a third point. Since the slope is m=4, it can be written as m=-4 over -1. Go back to (-1,-3) and count out the rise, -4, and the run, -1.

TRY IT 13.1

Graph the line with the point (-2,1) and slope m=3.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 1) and (negative 1, 4).

EXAMPLE 13.2

Graph the line with the point (4,-2) and slope m=-2.

Show answer
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (4, negative 2) and (5, negative 4).

Solve Slope Applications

At the beginning of this section, we said there are many applications of slope in the real world. Let’s look at a few now.

EXAMPLE 14

The ‘pitch’ of a building’s roof is the slope of the roof. Knowing the pitch is important in climates where there is heavy snowfall. If the roof is too flat, the weight of the snow may cause it to collapse. What is the slope of the roof shown?

This figure shows a house with a sloped roof. The roof on one half of the building is labeled "pitch of the roof". There is a line segment with arrows at each end measuring the vertical length of the roof and is labeled "rise equals 9 feet". There is a line segment with arrows at each end measuring the horizontal length of the root and is labeled "run equals 18 feet".

Solution
Use the slope formula. m=rise over run
Substitute the values for rise and run. m=9 over 18
Simplify. m=1 over 2
The slope of the roof is 1 over 2.
The roof rises 1 foot for every 2 feet of horizontal run.

TRY IT 14.1

Use (Example 14), substituting the rise = 14 and run = 24

Show answer

7 over 12

TRY IT 14.2

Use (Example 14), substituting rise = 15 and run = 36

Show answer

5 over 12

EXAMPLE 15

Have you ever thought about the sewage pipes going from your house to the street? They must slope down 1 over 4 inch per foot in order to drain properly. What is the required slope?

This figure is a right triangle. One leg is negative one quarter inch and the other leg is one foot.

Solution
Use the slope formula. mathematical expression
Simplify. m=-1 over 48
The slope of the pipe is -1 over 48.

The pipe drops 1 inch for every 48 inches of horizontal run.

TRY IT 15.1

Find the slope of a pipe that slopes down 1 over 3 inch per foot.

Show answer

-1 over 36

TRY IT 15.2

Find the slope of a pipe that slopes down 3 over 4 inch per yard.

Show answer

-1 over 48

Access these online resources for additional instruction and practice with understanding slope of a line.

Key Concepts

  • Find the Slope of a Line from its Graph using m=rise over run
    1. Locate two points on the line whose coordinates are integers.
    2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
    3. Count the rise and the run on the legs of the triangle.
    4. Take the ratio of rise to run to find the slope.
  • Graph a Line Given a Point and the Slope
    1. Plot the given point.
    2. Use the slope formula m=rise over run to identify the rise and the run.
    3. Starting at the given point, count out the rise and run to mark the second point.
    4. Connect the points with a line.
  • Slope of a Horizontal Line
    • The slope of a horizontal line, y=b, is 0.
  • Slope of a vertical line
    • The slope of a vertical line, x=a, is undefined

Glossary

geoboard
A geoboard is a board with a grid of pegs on it.
negative slope
A negative slope of a line goes down as you read from left to right.
positive slope
A positive slope of a line goes up as you read from left to right.
rise
The rise of a line is its vertical change.
run
The run of a line is its horizontal change.
slope formula
The slope of the line between two points (x sub 1,y sub 1) and (x sub 2,y sub 2) is m=y sub 2-y sub 1 over x sub 2-x sub 1.
slope of a line
The slope of a line is m=rise over run. The rise measures the vertical change and the run measures the horizontal change.

Practice Makes Perfect

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

1.
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 3 and the peg in column 5, row 2, forming a line.

2.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 4 and the peg in column 5, row 2, forming a line.
3.
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 4, row 2, forming a line.

4.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 3, row 4 and the peg in column 5, row 1, forming a line.
5.
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 1 and the peg in column 4, row 4, forming a line.

6.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 3 and the peg in column 5, row 4, forming a line.
7.
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 1 and the peg in column 5, row 4, forming a line.

8.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 2 and the peg in column 4, row 5, forming a line.

In the following exercises, model each slope. Draw a picture to show your results.

9. 2 over 3 10. 3 over 4
11. 1 over 4 12. 4 over 3
13. -1 over 2 14. -3 over 4
15. -2 over 3 16. -3 over 2

Use m=rise over run to find the Slope of a Line from its Graph

In the following exercises, find the slope of each line shown.

17.
The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line passes through the points (negative 10, negative 8), (0, negative 4), and (10, 0).

18.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, negative 8) and (2, negative 2).
19.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 4, negative 6) and (4, 4).

20.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line intercepts the y-axis at (0, negative 2) and passes through the point (3, 3).
21.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 3, 3) and (3, 1).

22.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 4) and (2, 2).
23.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line intercepts the y-axis at (0, 6) and passes through the point (4, 3).

24.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the point (negative 3, 1) and intercepts the y-axis at (0, negative 1).
25.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 1) and (2, 4).

26.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 1, 1) and (2, 3).
27.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 1, 6) and (1, 1).

28.

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the point (negative 1, 3) and intercepts the x-axis at (3, 0).
29.
The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 6) and (1, 4).

30.

The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line passes through the points (negative 1, 3) and (1, 2).
31.
The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line intercepts the x-axis at (negative 2, 0) and passes through the point (2, 1).

32.

The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line passes through the points (4, 2) and (7, 3).

Find the Slope of Horizontal and Vertical Lines

In the following exercises, find the slope of each line.

33. y=3 34. y=1
35. x=4 36. x=2
37. y=-2 38. y=-3
39. x=-5 40. x=-4

Use the Slope Formula to find the Slope of a Line between Two Points

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

41. (1,4),(3,9) 42. (2,3),(5,7)
43. (0,3),(4,6) 44. (0,1),(5,4)
45. (2,5),(4,0) 46. (3,6),(8,0)
47. (-3,3),(4,-5) 48. (-2,4),(3,-1)
49. (-1,-2),(2,5) 50. (-2,-1),(6,5)
51. (4,-5),(1,-2) 52. (3,-6),(2,-2)

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

53. (1,-2); m=3 over 4 54. (1,-1); m=2 over 3
55. (2,5); m=-1 over 3 56. (1,4); m=-1 over 2
57. (-3,4); m=-3 over 2 58. (-2,5); m=-5 over 4
59. (-1,-4); m=4 over 3 60.(-3,-5); m=3 over 2
61. y-intercept 3; m=-2 over 5 62. y-intercept 5; m=-4 over 3
63. x-intercept -2; m=3 over 4 64. x-intercept -1; m= 1 over 5
65. (-3,3); m=2 66. (-4,2); m=4
67. (1,5); m=-3 67. (1,5); m=-3

Everyday Math

69. Slope of a roof. An easy way to determine the slope of a roof is to set one end of a 12 inch level on the roof surface and hold it level. Then take a tape measure or ruler and measure from the other end of the level down to the roof surface. This will give you the slope of the roof. Builders, sometimes, refer to this as pitch and state it as an “x 12 pitch” meaning x over 12, where x is the measurement from the roof to the level—the rise. It is also sometimes stated as an “x-in-12 pitch”.

  1. a) What is the slope of the roof in this picture?
  2. b) What is the pitch in construction terms?
    This figure shows one side of a sloped roof of a house. The rise of the roof is labeled “4 inches” and the run of the roof is labeled “12 inches”.

70. The slope of the roof shown here is measured with a 12” level and a ruler. What is the slope of this roof?

This figure shows one side of a sloped roof of a house. The rise of the roof is measured with a ruler and shown to be 7 inches. The run of the roof is measured with a twelve inch level and shown to be 12 inches.

71. Road grade. A local road has a grade of 6%. The grade of a road is its slope expressed as a percent. Find the slope of the road as a fraction and then simplify. What rise and run would reflect this slope or grade?

72. Highway grade. A local road rises 2 feet for every 50 feet of highway.

a) What is the slope of the highway?
b) The grade of a highway is its slope expressed as a percent. What is the grade of this highway?

73. Wheelchair ramp. The rules for wheelchair ramps require a maximum 1-inch rise for a 12-inch run.

a) How long must the ramp be to accommodate a 24-inch rise to the door?
b) Create a model of this ramp.

74. Wheelchair ramp. A 1-inch rise for a 16-inch run makes it easier for the wheelchair rider to ascend a ramp.

a) How long must a ramp be to easily accommodate a 24-inch rise to the door?
b) Create a model of this ramp.

Writing Exercises

75. What does the sign of the slope tell you about a line? 76. How does the graph of a line with slope m=1 over 2 differ from the graph of a line with slope m=2?
77. Why is the slope of a vertical line “undefined”?

Answers

 

1. 1 over 4 3. 2 over 3
5. -3 over 2=-3 over 2 7. -3 over 4
9.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 5 and the peg in column 5, row 3, forming a line.

11.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 5, row 3, forming a line.

13.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 3, row 5, forming a line.

15.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 2 and the peg in column 4, row 4, forming a line.

17. 2 over 5 19. 5 over 4
21. -1 over 3 23. -3 over 4
25. 3 over 4 27. -5 over 2
29. -2 over 3 31. 1 over 4
33. 0 35. undefined
37. 0 39. undefined
41. 5 over 2 43. 3 over 4
45. -5 over 2 47. -8 over 7
49. 7 over 3 51. -1
53.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (1, negative 2) and (5, 1).

55.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (2, 5) and (5, 4).

57.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 3, 4) and (negative 1, 1).

59.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 1, negative 4) and intercepts the x-axis at (2, 0).

61.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the y-axis at (0, 3) and passes through the point (5, 1).

63.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the x-axis at (negative 2, 0) and passes through the point (2, 3).

65.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 3, 3) and (negative 2, 5).

67.

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (1, 5) and (2, 2).

69. a)1 over 3 b) 4 12 pitch or 4-in-12 pitch 71. 3 over 50; rise = 3, run = 50
73. a) 288 inches (24 feet) b) Models will vary. 75. When the slope is a positive number the line goes up from left to right. When the slope is a negative number the line goes down from left to right.
77. A vertical line has 0 run and since division by 0 is undefined the slope is undefined.

Attributions

This chapter has been adapted from “Understand Slope of a Line” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

36

6.5 Use the Slope–Intercept Form of an Equation of a Line

Learning Objectives

By the end of this section, you will be able to:

  • Recognize the relation between the graph and the slope–intercept form of an equation of a line
  • Identify the slope and y-intercept form of an equation of a line
  • Graph a line using its slope and intercept
  • Choose the most convenient method to graph a line
  • Graph and interpret applications of slope–intercept
  • Use slopes to identify parallel lines
  • Use slopes to identify perpendicular lines

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using the point–slope method. Once we see how an equation in slope–intercept form and its graph are related, we’ll have one more method we can use to graph lines.

In Graph Linear Equations in Two Variables, we graphed the line of the equation y=1 over 2x+3 by plotting points. See (Figure). Let’s find the slope of this line.

This figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line is labeled with the equation y equals one half x, plus 3. The points (0, 3), (2, 4) and (4, 5) are labeled also. A red vertical line begins at the point (2, 4) and ends one unit above the point. It is labeled “Rise equals 1”. A red horizontal line begins at the end of the vertical line and ends at the point (4, 5). It is labeled “Run equals 2. The red lines create a right triangle with the line y equals one half x, plus 3 as the hypotenuse.

The red lines show us the rise is 1 and the run is 2. Substituting into the slope formula:

mathematical expression

What is the y-intercept of the line? The y-intercept is where the line crosses the y-axis, so y-intercept is (0,3). The equation of this line is:

The figure shows the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue.

Notice, the line has:

The figure shows the statement “slope m equals one half and y-intercept (0, 3). The slope, one half, is colored red and the number 3 in the y-intercept is colored blue.

When a linear equation is solved for y, the coefficient of the x term is the slope and the constant term is the y-coordinate of the y-intercept. We say that the equation y=1 over 2x+3 is in slope–intercept form.

The figure shows the statement “m equals one half; y-intercept is (0, 3). The slope, one half, is colored red and the number 3 in the y-intercept is colored blue. Below that statement is the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue. Below the equation is another equation y equals m x, plus b. The variable m is colored red and the variable b is colored blue.

Slope-intercept form of an equation of a line

The slope–intercept form of an equation of a line with slope m and y-intercept, (0,b) is,

y=mx+b

Sometimes the slope–intercept form is called the “y-form.”

EXAMPLE 1

Use the graph to find the slope and y-intercept of the line, y=2x+1.

Compare these values to the equationy=mx+b.

Solution

To find the slope of the line, we need to choose two points on the line. We’ll use the points (0,1) and (1,3).

.
Find the rise and run. .
.
.
Find the y-intercept of the line. The y-intercept is the point (0, 1).
. .

The slope is the same as the coefficient of x and the y-coordinate of the y-intercept is the same as the constant term.

TRY IT 1.1

Use the graph to find the slope and y-intercept of the line y=2 over 3x-1. Compare these values to the equation y=mx+b.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line goes through the points (0, negative 1) and (6, 3).

Show answer

slope m=2 over 3 and y-intercept (0,-1)

TRY IT 1.2

Use the graph to find the slope and y-intercept of the line y=1 over 2x+3. Compare these values to the equation y=mx+b.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line goes through the points (0, 3) and (negative 6, 0).

Show answer

slope m=1 over 2 and y-intercept (0,3)

Identify the Slope and y-Intercept From an Equation of a Line

In Understand Slope of a Line, we graphed a line using the slope and a point. When we are given an equation in slope–intercept form, we can use the y-intercept as the point, and then count out the slope from there. Let’s practice finding the values of the slope and y-intercept from the equation of a line.

EXAMPLE 2

Identify the slope and y-intercept of the line with equation y=-3x+5.

Solution

We compare our equation to the slope–intercept form of the equation.

.
Write the equation of the line. .
Identify the slope. .
Identify the y-intercept. .

TRY IT 2.1

Identify the slope and y-intercept of the line y=2 over 5x-1.

Show answer

2 over 5;(0,-1)

TRY IT 2.2

Identify the slope and y-intercept of the line y=-4 over 3x+1.

Show answer

-4 over 3;(0,1)

When an equation of a line is not given in slope–intercept form, our first step will be to solve the equation for y.

EXAMPLE 3

Identify the slope and y-intercept of the line with equation x+2y=6.

Solution

This equation is not in slope–intercept form. In order to compare it to the slope–intercept form we must first solve the equation fory.

Solve for y. x+2y=6
Subtract x from each side. .
Divide both sides by 2. .
Simplify. .
(Remember:a+b over c=a over c+b over c)
Simplify. .
Write the slope–intercept form of the equation of the line. .
Write the equation of the line. .
Identify the slope. .
Identify the y-intercept. .

TRY IT 3.1

Identify the slope and y-intercept of the line x+4y=8.

Show answer

-1 over 4;(0,2)

TRY IT 3.2

Identify the slope and y-intercept of the line 3x+2y=12.

Show answer

-3 over 2;(0,6)

Graph a Line Using its Slope and Intercept

Now that we know how to find the slope and y-intercept of a line from its equation, we can graph the line by plotting the y-intercept and then using the slope to find another point.

EXAMPLE 4

How to Graph a Line Using its Slope and Intercept

Graph the line of the equation y=4x-2 using its slope and y-intercept.

Solution

The figure shows the steps to graph the equation y equals 4x minus 2. Step 1 is to find the slope intercept form of the equation. The equation is already in slope intercept form.Step 2 is to identify the slope and y-intercept. Use the equation y equals m x, plus b. The equation y equals m x, plus b is shown with the variable m colored red and the variable b colored blue. Below that is the equation y equals 4 x, plus -2. The number 4 is colored red and -2 is colored blue. From this equation we can see that m equals 4 and b equals -2 so the slope is 4 and the y-intercept is the point (0, negative 2).Step 3 is to plot the y-intercept. An x y-coordinate plane is shown with the x-axis of the plane running from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The point (0, negative 2) is plotted.Step 4 is to use the slope formula m equals rise over run to identify the rise and the run. Since m equals 4, rise over run equals 4 over 1. From this we can determine that the rise is 4 and the run is 1.Step 5 is to start at they-intercept, count out the rise and run to mark the second point. So start at the point (0, negative 2) and count the rise and the run. The rise is up 4 and the run is right 1. On the x y-coordinate plane is a red vertical line starts at the point (0, negative 2) and rises 4 units at its end a red horizontal line runs 1 unit to end at the point (1, 2). The point (1, 2) is plotted.Step 6 is to connect the points with a line. On the x y-coordinate plane the points (0, negative 2) and (1, 2) are plotted and a line runs through the two points. The line is the graph of y equals 4 x, minus 2.

TRY IT 4.1

Graph the line of the equation y=4x+1 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, 1) and (1, 5) are plotted on the line.

TRY IT 4.2

Graph the line of the equation y=2x-3 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 3) and (1, negative 1) are plotted on the line.

HOW TO: Graph a line using its slope and y-intercept

  1. Find the slope-intercept form of the equation of the line.
  2. Identify the slope and y-intercept.
  3. Plot the y-intercept.
  4. Use the slope formula m=rise over run to identify the rise and the run.
  5. Starting at the y-intercept, count out the rise and run to mark the second point.
  6. Connect the points with a line.

EXAMPLE 5

Graph the line of the equation y=-x+4 using its slope and y-intercept.

Solution
y=mx+b
The equation is in slope–intercept form. y=-x+4
Identify the slope and y-intercept. m=-1
y-intercept is (0, 4)
Plot the y-intercept. See graph below.
Identify the rise and the run. m=-1 over 1
Count out the rise and run to mark the second point. rise −1, run 1
Draw the line. .
To check your work, you can find another point on the line and make sure it is a solution of the equation. In the graph we see the line goes through (4, 0).
Check.
mathematical expression

TRY IT 5.1

Graph the line of the equation y=-x-3 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 3) and (1, negative 4) are plotted on the line.

TRY IT 5.2

Graph the line of the equation y=-x-1 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 1) and (1, negative 2) are plotted on the line.

EXAMPLE 6

Graph the line of the equation y=-2 over 3x-3 using its slope and y-intercept.


Solution
y=mx+b
The equation is in slope–intercept form. y=-2 over 3x-3
Identify the slope and y-intercept. m=-2 over 3; y-intercept is (0, −3)
Plot the y-intercept. See graph below.
Identify the rise and the run.
Count out the rise and run to mark the second point.
Draw the line. .

TRY IT 6.1

Graph the line of the equation y=-5 over 2x+1 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0,1) and (2, negative 4) are plotted on the line.

TRY IT 6.2

Graph the line of the equation y=-3 over 4x-2 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 2) and (4, negative 5) are plotted on the line.

EXAMPLE 7

Graph the line of the equation 4x-3y=12 using its slope and y-intercept.

Solution
4x-3y=12
Find the slope–intercept form of the equation. -3y=-4x+12
-3y over 3=-4x+12 over -3
The equation is now in slope–intercept form. y=4 over 3x-4
Identify the slope and y-intercept. m=4 over 3
y-intercept is (0, −4)
Plot the y-intercept. See graph below.
Identify the rise and the run; count out the rise and run to mark the second point.
Draw the line. .

TRY IT 7.1

Graph the line of the equation 2x-y=6 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 6) and (1, negative 4) are plotted on the line.

TRY IT 7.2

Graph the line of the equation 3x-2y=8 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 4) and (2, negative 1) are plotted on the line.

We have used a grid with x and y both going from about -10 to 10 for all the equations we’ve graphed so far. Not all linear equations can be graphed on this small grid. Often, especially in applications with real-world data, we’ll need to extend the axes to bigger positive or smaller negative numbers.

EXAMPLE 8

Graph the line of the equation y=0.2x+45 using its slope and y-intercept.


Solution

We’ll use a grid with the axes going from about -80 to 80.

y=mx+b
The equation is in slope–intercept form. y=0.2x+45
Identify the slope and y-intercept. m=0.2
The y-intercept is (0, 45)
Plot the y-intercept. See graph below.
Count out the rise and run to mark the second point. The slope is m=0.2; in fraction form this means m=2 over 10. Given the scale of our graph, it would be easier to use the equivalent fraction m=10 over 50.
Draw the line. .

TRY IT 8.1

Graph the line of the equation y=0.5x+25 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 70 to 30. The y-axis of the plane runs from negative 20 to 40. The points (0, 25) and (10, 30) are plotted on the line.

TRY IT 8.2

Graph the line of the equation y=0.1x-30 using its slope and y-intercept.

Show answer
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 50 to 350. The y-axis of the plane runs from negative 40 to 40. The points (0, negative 30) and (100, negative 20) are plotted on the line.
Now that we have graphed lines by using the slope and y-intercept, let’s summarize all the methods we have used to graph lines. See (Figure).

Methods to graph lines

The table has two rows and four columns. The first row spans all four columns and is a header row. The header is “Methods to Graph Lines”. The second row is made up of four columns. The first column is labeled “Plotting Points” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The rest of the table is blank. Below the table it reads “Find three points. Plot the points, make sure they line up, then draw the line.” The Second column is labeled “Slope–Intercept” and shows the equation y equals m x, plus b. Below the equation it reads “Find the slope and y-intercept. Start at the y-intercept, then count the slope to get a second point.” The third column is labeled “Intercepts” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The second row has a 0 in the “x” column and the “y” column is blank. The second row is blank in the “x” column and has a 0 in the “y” column. The third row is blank. Below the table it reads “Find the intercepts and a third point. Plot the points, make sure they line up, then draw the line.” The fourth column is labeled “Recognize Vertical and Horizontal Lines”. Below that it reads “The equation has only one variable.” The equation x equals a is a vertical line and the equation y equals b is a horizontal line.

Choose the Most Convenient Method to Graph a Line

Now that we have seen several methods we can use to graph lines, how do we know which method to use for a given equation?

While we could plot points, use the slope–intercept form, or find the intercepts for any equation, if we recognize the most convenient way to graph a certain type of equation, our work will be easier. Generally, plotting points is not the most efficient way to graph a line. We saw better methods in sections 4.3, 4.4, and earlier in this section. Let’s look for some patterns to help determine the most convenient method to graph a line.

Here are six equations we graphed in this chapter, and the method we used to graph each of them.

mathematical expression

Equations #1 and #2 each have just one variable. Remember, in equations of this form the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations #3 and #4, both x and y are on the same side of the equation. These two equations are of the form Ax+By=C. We substituted y=0 to find the x-intercept and x=0 to find the y-intercept, and then found a third point by choosing another value for x or y.

Equations #5 and #6 are written in slope–intercept form. After identifying the slope and y-intercept from the equation we used them to graph the line.

This leads to the following strategy.

Strategy for choosing the most convenient method to graph a line

Consider the form of the equation.

  • If it only has one variable, it is a vertical or horizontal line.
    • x=a is a vertical line passing through the x-axis at a.
    • y=b is a horizontal line passing through the y-axis at b.
  • If y is isolated on one side of the equation, in the form y=mx+b, graph by using the slope and y-intercept.
    • Identify the slope and y-intercept and then graph.
  • If the equation is of the form Ax+By=C, find the intercepts.
    • Find the x– and y-intercepts, a third point, and then graph.

EXAMPLE 9

Determine the most convenient method to graph each line.

a) y=-6 b )5x-3y=15 c) x=7 d) y=2 over 5x-1.

Solution
  1. y=-6
    This equation has only one variable,y. Its graph is a horizontal line crossing the y-axis at -6.
  2. 5x-3y=15
    This equation is of the form Ax+By=C. The easiest way to graph it will be to find the intercepts and one more point.
  3. x=7
    There is only one variable, x. The graph is a vertical line crossing the x-axis at 7.
  4. y=2 over 5x-1
    Since this equation is in y=mx+b form, it will be easiest to graph this line by using the slope and y-intercept.

TRY IT 9.1

Determine the most convenient method to graph each line: a) 3x+2y=12 b) y=4 c) y=1 over 5x-4 d) x=-7.

Show answer

a) intercepts b) horizontal line c) slope–intercept d) vertical line

TRY IT 9.2

Determine the most convenient method to graph each line: a) x=6 b) y=-3 over 4x+1 c) y=-8 d) 4x-3y=-1.

Show answer

a) vertical line b) slope–intercept c) horizontal line d) intercepts

Graph and Interpret Applications of Slope–Intercept

Many real-world applications are modeled by linear equations. We will take a look at a few applications here so you can see how equations written in slope–intercept form relate to real-world situations.

Usually when a linear equation models a real-world situation, different letters are used for the variables, instead of x and y. The variable names remind us of what quantities are being measured.

EXAMPLE 10

The equation F=9 over 5C+32 is used to convert temperatures, C, on the Celsius scale to temperatures, F, on the Fahrenheit scale.

a) Find the Fahrenheit temperature for a Celsius temperature of 0.
b) Find the Fahrenheit temperature for a Celsius temperature of 20.
c) Interpret the slope and F-intercept of the equation.
d) Graph the equation.

 

Solution
a)
Find the Fahrenheit temperature for a Celsius temperature of 0.
Find F when C=0.
Simplify.
mathematical expression
b)
Find the Fahrenheit temperature for a Celsius temperature of 20.
Find F when C=20.
Simplify.
Simplify.
mathematical expression

c) Interpret the slope and F-intercept of the equation.

Even though this equation uses Fand C, it is still in slope–intercept form.

This image shows three lines of equations. The first line reads y equals m x plus b. The second line reads F equals m C plus b and the third line reads F equals nine fifths times C plus 32.

The slope, 9 over 5, means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) increases 5 degrees.

The F-intercept means that when the temperature is 0° on the Celsius scale, it is 32° on the Fahrenheit scale.

d) Graph the equation.

We’ll need to use a larger scale than our usual. Start at the F-intercept (0,32) then count out the rise of 9 and the run of 5 to get a second point. See (Figure).

No Alt Text

TRY IT 10.1

The equation h=2s+50 is used to estimate a woman’s height in inches, h, based on her shoe size, s.

a) Estimate the height of a child who wears women’s shoe size 0.
b) Estimate the height of a woman with shoe size 8.
c) Interpret the slope and h-intercept of the equation.
d) Graph the equation.

Show answer
  1. 50 inches
  2. 66 inches
  3. The slope, 2, means that the height, h, increases by 2 inches when the shoe size, s, increases by 1. The h-intercept means that when the shoe size is 0, the height is 50 inches.
  4. The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable s and runs from negative 2 to 15. The y-axis of the plane represents the variable h and runs from negative 1 to 80. The line begins at the point (0, 50) and goes through the points (8, 66).

TRY IT 10.2

The equation T=1 over 4n+40 is used to estimate the temperature in degrees Fahrenheit, T, based on the number of cricket chirps, n, in one minute.

a) Estimate the temperature when there are no chirps.
b) Estimate the temperature when the number of chirps in one minute is 100.
c) Interpret the slope and T-intercept of the equation.
d) Graph the equation.

Show answer
  1. 40 degrees
  2. 65 degrees
  3. The slope, 1 over 4, means that the temperature Fahrenheit (F) increases 1 degree when the number of chirps, n, increases by 4. The T-intercept means that when the number of chirps is 0, the temperature is 40°.
  4. The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable n and runs from 10 to 140 The y-axis of the plane represents the variable T and runs from negative 5 to 75. The line begins at the point (0, 40) and goes through the point (100, 65).

The cost of running some types business has two components—a fixed cost and a variable cost. The fixed cost is always the same regardless of how many units are produced. This is the cost of rent, insurance, equipment, advertising, and other items that must be paid regularly. The variable cost depends on the number of units produced. It is for the material and labour needed to produce each item.

EXAMPLE 11

Stella has a home business selling gourmet pizzas. The equation C=4p+25 models the relation between her weekly cost, C, in dollars and the number of pizzas, p, that she sells.

a) Find Stella’s cost for a week when she sells no pizzas.
b) Find the cost for a week when she sells 15 pizzas.
c) Interpret the slope and C-intercept of the equation.
d) Graph the equation.

Solution
a) Find Stella’s cost for a week when she sells no pizzas. .
Find C when p=0. .
Simplify. .
Stella’s fixed cost is $25 when she sells no pizzas.
b) Find the cost for a week when she sells 15 pizzas. .
Find C when p=15. .
Simplify. .
.
Stella’s costs are $85 when she sells 15 pizzas.
c) Interpret the slope and C-intercept of the equation. .
The slope, 4, means that the cost increases by $4 for each pizza Stella sells. The C-intercept means that even when Stella sells no pizzas, her costs for the week are $25.
d) Graph the equation. We’ll need to use a larger scale than our usual. Start at the C-intercept (0, 25) then count out the rise of 4 and the run of 1 to get a second point. .

TRY IT 11.1

Sam drives a delivery van. The equation C=0.5m+60 models the relation between his weekly cost, C, in dollars and the number of miles, m, that he drives.

a) Find Sam’s cost for a week when he drives 0 miles.
b) Find the cost for a week when he drives 250 miles.
c) Interpret the slope and C-intercept of the equation.
d) Graph the equation.

Show answer
  1. $60
  2. $185
  3. The slope, 0.5, means that the weekly cost, C, increases by $0.50 when the number of miles driven, n, increases by 1. The C-intercept means that when the number of miles driven is 0, the weekly cost is $60
  4. The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable m and runs from negative 10 to 400. The y-axis of the plane represents the variable C and runs from negative 10 to 300. The line begins at the point (0, 65) and goes through the point (250, 185).

TRY IT 11.2

Loreen has a calligraphy business. The equation C=1.8n+35 models the relation between her weekly cost, C, in dollars and the number of wedding invitations, n, that she writes.

a) Find Loreen’s cost for a week when she writes no invitations.
b) Find the cost for a week when she writes 75 invitations.
c) Interpret the slope and C-intercept of the equation.
d) Graph the equation.

Show answer
  1. $35
  2. $170
  3. The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n, increases by 1.80.
    The C-intercept means that when the number of invitations is 0, the weekly cost is $35.;
  4. The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable n and runs from negative 10 to 400. The y-axis of the plane represents the variable C and runs from negative 10 to 300. The line begins at the point (0, 35) and goes through the point (75, 170).

Use Slopes to Identify Parallel Lines

The slope of a line indicates how steep the line is and whether it rises or falls as we read it from left to right. Two lines that have the same slope are called parallel lines. Parallel lines never intersect.

The figure shows three pairs of lines side-by-side. The pair of lines on the left run diagonally rising from left to right. The pair run side-by-side, not crossing. The pair of lines in the middle run diagonally dropping from left to right. The pair run side-by-side, not crossing. The pair of lines on the right run diagonally also dropping from left to right, but with a lesser slope. The pair run side-by-side, not crossing.

We say this more formally in terms of the rectangular coordinate system. Two lines that have the same slope and different y-intercepts are called parallel lines. See (Figure).

Verify that both lines have the same slope, m=2 over 5, and different y-intercepts.

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (negative 5,1) and (5,5). The other line goes through the points (negative 5, negative 4) and (5,0).

What about vertical lines? The slope of a vertical line is undefined, so vertical lines don’t fit in the definition above. We say that vertical lines that have different x-intercepts are parallel. See (Figure).

Vertical lines with different x-intercepts are parallel.

The figure shows two vertical lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (2,1) and (2,5). The other line goes through the points (5, negative 4) and (5,0).

Parallel lines

Parallel lines are lines in the same plane that do not intersect.

  • Parallel lines have the same slope and different y-intercepts.
  • If m sub 1 and m sub 2 are the slopes of two parallel lines thenm sub 1=m sub 2.
  • Parallel vertical lines have different x-intercepts.

Let’s graph the equations y=-2x+3 and 2x+y=-1 on the same grid. The first equation is already in slope–intercept form: y=-2x+3. We solve the second equation for y:

mathematical expression

Graph the lines.

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (negative 4, 7) and (3, negative 7). The other line goes through the points (negative 2, 7) and (5, negative 7).

Notice the lines look parallel. What is the slope of each line? What is the y-intercept of each line?

mathematical expression

The slopes of the lines are the same and the y-intercept of each line is different. So we know these lines are parallel.

Since parallel lines have the same slope and different y-intercepts, we can now just look at the slope–intercept form of the equations of lines and decide if the lines are parallel.

EXAMPLE 12

Use slopes and y-intercepts to determine if the lines 3x-2y=6 and y=3 over 2x+1 are parallel.

Solution
Solve the first equation for y. mathematical expression and y=3 over 2x+1
The equation is now in slope-intercept form. y=3 over 2x-3
The equation of the second line is already in slope-intercept form. y=3 over 2x+1
Identify the slope and y-intercept of both lines. mathematical expression mathematical expression
y-intercept is (0, −3) y-intercept is (0, 1)

The lines have the same slope and different y-intercepts and so they are parallel. You may want to graph the lines to confirm whether they are parallel.

TRY IT 12.1

Use slopes and y-intercepts to determine if the lines 2x+5y=5 and y=-2 over 5x-4 are parallel.

Show answer

parallel

TRY IT 12.2

Use slopes and y-intercepts to determine if the lines 4x-3y=6 and y=4 over 3x-1 are parallel.

Show answer

parallel

EXAMPLE 13

Use slopes and y-intercepts to determine if the lines y=-4 and y=3 are parallel.

Solution
mathematical expression and mathematical expression
Write each equation in slope-intercept form. y=0x-4 y=0x+3
Since there is no x term we write 0x. y=mx+b y=mx+b
Identify the slope and y-intercept of both lines. m=0 m=0
y-intercept is (0, 4) y-intercept is (0, 3)

The lines have the same slope and different y-intercepts and so they are parallel.

There is another way you can look at this example. If you recognize right away from the equations that these are horizontal lines, you know their slopes are both 0. Since the horizontal lines cross the y-axis at y=-4 and at y=3, we know the y-intercepts are (0,-4) and (0,3). The lines have the same slope and different y-intercepts and so they are parallel.

TRY IT 13.1

Use slopes and y-intercepts to determine if the lines y=8 and y=-6 are parallel.

Show answer

parallel

TRY IT 13.2

Use slopes and y-intercepts to determine if the lines y=1 and y=-5 are parallel.

Show answer

parallel

EXAMPLE 14

Use slopes and y-intercepts to determine if the lines x=-2 and x=-5 are parallel.

Solution
x=-2 and x=-5

Since there is noy, the equations cannot be put in slope–intercept form. But we recognize them as equations of vertical lines. Their x-intercepts are -2 and -5. Since their x-intercepts are different, the vertical lines are parallel.

TRY IT 14.1

Use slopes and y-intercepts to determine if the lines x=1 and x=-5 are parallel.

Show answer

parallel

TRY IT 14.2

Use slopes and y-intercepts to determine if the lines x=8 and x=-6 are parallel.

Show answer

parallel

EXAMPLE 15

Use slopes and y-intercepts to determine if the lines y=2x-3 and -6x+3y=-9 are parallel. You may want to graph these lines, too, to see what they look like.

Solution
y=2x-3 and -6x+3y=-9
The first equation is already in slope-intercept form. y=2x-3
Solve the second equation for y. mathematical expression
The second equation is now in slope-intercept form. y=2x-3
Identify the slope and y-intercept of both lines. mathematical expression mathematical expression
y-intercept is (0, -3) y-intercept is (0, -3)

The lines have the same slope, but they also have the same y-intercepts. Their equations represent the same line. They are not parallel; they are the same line.

TRY IT 15.1

Use slopes and y-intercepts to determine if the lines y=-1 over 2x-1 and x+2y=2 are parallel.

Show answer

not parallel; same line

TRY IT 15.2

Use slopes and y-intercepts to determine if the lines y=3 over 4x-3 and 3x-4y=12 are parallel.

Show answer

not parallel; same line

Use Slopes to Identify Perpendicular Lines

Let’s look at the lines whose equations are y=1 over 4x-1 and y=-4x+2, shown in (Figure).

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line is labeled with the equation y equals negative 4x plus 2 and goes through the points (0,2) and (1, negative 2). The other line is labeled with the equation y equals one fourth x minus 1 and goes through the points (0, negative 1) and (4,0).

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.

What do you notice about the slopes of these two lines? As we read from left to right, the line y=1 over 4x-1 rises, so its slope is positive. The liney=-4x+2 drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, m sub 1=1 over 4, and the slope of the second line, m sub 2=-4, we can see that they are negative reciprocals of each other. If we multiply them, their product is -1.

mathematical expression

This is always true for perpendicular lines and leads us to this definition.

Perpendicular lines

Perpendicular lines are lines in the same plane that form a right angle.

If m sub 1 and m sub 2 are the slopes of two perpendicular lines, then:

m sub 1 times m sub 2=-1 and m sub 1=-1 over m sub 2

Vertical lines and horizontal lines are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is -1, the lines are perpendicular. Perpendicular lines may have the same y-intercepts.

EXAMPLE 16

Use slopes to determine if the lines, y=-5x-4 and x-5y=5 are perpendicular.

Solution
The first equation is already in slope-intercept form. y = -5x-4
Solve the second equation for y. mathematical expression
Identify the slope of each line. mathematical expression mathematical expression

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

mathematical expression

TRY IT 16.1

Use slopes to determine if the lines y=-3x+2 and x-3y=4 are perpendicular.

Show answer

perpendicular

TRY IT 16.2

Use slopes to determine if the lines y=2x-5 and x+2y=-6 are perpendicular.

Show answer

perpendicular

EXAMPLE 17

Use slopes to determine if the lines, 7x+2y=3 and 2x+7y=5 are perpendicular.

Solution
Solve the equations for y. mathematical expression mathematical expression
Identify the slope of each line. mathematical expression mathematical expression

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

TRY IT 17.1

Use slopes to determine if the lines 5x+4y=1 and 4x+5y=3 are perpendicular.

Show answer

not perpendicular

TRY IT 17.2

Use slopes to determine if the lines 2x-9y=3 and 9x-2y=1 are perpendicular.

Show answer

not perpendicular

Access this online resource for additional instruction and practice with graphs.

Key Concepts

  • The slope–intercept form of an equation of a line with slope m and y-intercept, (0,b) is, y=mx+b.
  • Graph a Line Using its Slope and y-Intercept
    1. Find the slope-intercept form of the equation of the line.
    2. Identify the slope and y-intercept.
    3. Plot the y-intercept.
    4. Use the slope formula m=rise over run to identify the rise and the run.
    5. Starting at the y-intercept, count out the rise and run to mark the second point.
    6. Connect the points with a line.
  • Strategy for Choosing the Most Convenient Method to Graph a Line: Consider the form of the equation.
    • If it only has one variable, it is a vertical or horizontal line.
      x=a is a vertical line passing through the x-axis at a.
      y=b is a horizontal line passing through the y-axis at b.
    • If y is isolated on one side of the equation, in the form y=mx+b, graph by using the slope and y-intercept.
      Identify the slope and y-intercept and then graph.
    • If the equation is of the form Ax+By=C, find the intercepts.
      Find the x– and y-intercepts, a third point, and then graph.
  • Parallel lines are lines in the same plane that do not intersect.
    • Parallel lines have the same slope and different y-intercepts.
    • If m1 and m2 are the slopes of two parallel lines then m sub 1=m sub 2.
    • Parallel vertical lines have different x-intercepts.
  • Perpendicular lines are lines in the same plane that form a right angle.
    • If mathematical expression are the slopes of two perpendicular lines, then m sub 1 times m sub 2=-1 and m sub 1=-1 over m sub 2.
    • Vertical lines and horizontal lines are always perpendicular to each other.

Glossary

parallel lines
Lines in the same plane that do not intersect.
perpendicular lines
Lines in the same plane that form a right angle.
slope-intercept form of an equation of a line
The slope–intercept form of an equation of a line with slope m and y-intercept, (0,b) is, y=mx+b.

Practice Makes Perfect

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

In the following exercises, use the graph to find the slope and y-intercept of each line. Compare the values to the equation y=mx+b.

1.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 2) and (1,2).

y=4x-2

2.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 5) and (1, negative 2).

y=3x-5

3.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,1) and (1, negative 2).

y=-3x+1

4.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,4) and (1,3).

y=-x+4

5.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,3) and (1,5).

y=-2 over 5x+3

6.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,1) and (3, negative 3).

y=-4 over 3x+1

Identify the Slope and y-Intercept From an Equation of a Line

In the following exercises, identify the slope and y-intercept of each line.

7. y=-9x+7 8. y=-7x+3
9. y=4x-10 10.y=6x-8
11. 4x+y=8 12. 3x+y=5
13. 8x+3y=12 14. 6x+4y=12
15. 7x-3y=9 16. 5x-2y=6

Graph a Line Using Its Slope and Intercept

In the following exercises, graph the line of each equation using its slope and y-intercept.

17. y=x+4 18. y=x+3
19. y=2x-3 20. y=3x-1
21. y=-x+3 22. y=-x+2
23. y=-x-2 24. y=-x-4
25. y=-2 over 5-3 26. y=-3 over 4-1
27. y=-2 over 3+1 28. y=-3 over 5+2
29. 4x-3y=6 30. 3x-4y=8
31. y=0.1x+15 32. y=0.1x+15

Choose the Most Convenient Method to Graph a Line

In the following exercises, determine the most convenient method to graph each line.

33. y=4 34. x=2
35. x=-3 36. y=5
37. y=-3x+4 38. y=-3x+4
39. x-y=1 40. x-y=5
41. y=4 over 5x-3 42. y=2 over 3x-1
43. y=-1 44. y=-3
45. 2x-5y=-10 46. 3x-2y=-12
47. y=-1 over 3x+5 48. y=-1 over 4+3

Graph and Interpret Applications of Slope–Intercept

49. The equation P=28+2.54w models the relation between the amount of Randy’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

  1. Find the payment for a month when Randy used 0 units of water.
  2. Find the payment for a month when Randy used 15 units of water.
  3. Interpret the slope and P-intercept of the equation.
  4. Graph the equation.

50. The equation P=31+1.75w models the relation between the amount of Tuyet’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

  1. Find Tuyet’s payment for a month when 0 units of water are used.
  2. Find Tuyet’s payment for a month when 12 units of water are used.
  3. Interpret the slope and P-intercept of the equation.
  4. Graph the equation.

51. Janelle is planning to rent a car while on vacation. The equation C=0.32m+15 models the relation between the cost in dollars, C, per day and the number of miles, m, she drives in one day.

  1. Find the cost if Janelle drives the car 0 miles one day.
  2. Find the cost on a day when Janelle drives the car 400 miles.
  3. Interpret the slope and C–intercept of the equation.
  4. Graph the equation.

52. Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day.

  1. Find the amount Bruce is reimbursed on a day when he drives 0 miles.
  2. Find the amount Bruce is reimbursed on a day when he drives 220 miles.
  3. Interpret the slope and R-intercept of the equation.
  4. Graph the equation.

53. Patel’s weekly salary includes a base pay plus commission on his sales. The equation S=750+0.09c models the relation between his weekly salary, S, in dollars and the amount of his sales, c, in dollars.

  1. Find Patel’s salary for a week when his sales were 0.
  2. Find Patel’s salary for a week when his sales were 18,540.
  3. Interpret the slope and S-intercept of the equation.
  4. Graph the equation.

54. Cherie works in retail and her weekly salary includes commission for the amount she sells. The equation S=400+0.15c models the relation between her weekly salary, S, in dollars and the amount of her sales, c, in dollars.

  1. Find Cherie’s salary for a week when her sales were 0.
  2. Find Cherie’s salary for a week when her sales were 3600.
  3. Interpret the slope and S–intercept of the equation.
  4. Graph the equation.

55. Margie is planning a dinner banquet. The equation C=750+42g models the relation between the cost in dollars, C of the banquet and the number of guests, g.

  1. Find the cost if the number of guests is 50.
  2. Find the cost if the number of guests is 100.
  3. Interpret the slope and C–intercept of the equation.
  4. Graph the equation.

56. Costa is planning a lunch banquet. The equation C=450+28g models the relation between the cost in dollars, C, of the banquet and the number of guests, g.

  1. Find the cost if the number of guests is 40.
  2. Find the cost if the number of guests is 80.
  3. Interpret the slope and C-intercept of the equation.
  4. Graph the equation.

Use Slopes to Identify Parallel Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel.

57. mathematical expression 58. mathematical expression
59. mathematical expression 60. mathematical expression
61. mathematical expression 62. mathematical expression
63. mathematical expression 64. mathematical expression
65. mathematical expression 66. mathematical expression
67. mathematical expression 68. mathematical expression
69. mathematical expression 70. mathematical expression
71. mathematical expression 72. mathematical expression
73. mathematical expression 74. mathematical expression
75. mathematical expression 76. mathematical expression
77. mathematical expression 78. mathematical expression
79. mathematical expression 80. mathematical expression
81. mathematical expression 82. mathematical expression

Use Slopes to Identify Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are perpendicular.

83. x-4y=8;4x+y=2 84. 3x-2y=8;2x+3y=6
85. 2x+3y=5;3x-2y=7 86. 2x+5y=3;5x-2y=6
87. 3x-4y=8;4x-3y=6 88. 3x-2y=1;2x-3y=2
89. 2x+4y=3;6x+3y=2 90. 5x+2y=6;2x+5y=8
91. 2x-6y=4;12x+4y=9 92. 4x-2y=5;3x+6y=8
93. 8x-2y=7;3x+12y=9 94. 6x-4y=5;8x+12y=3

Everyday Math

95. The equation n=4T-160 is used to estimate the number of cricket chirps, n, in one minute based on the temperature in degrees Fahrenheit, T.

  1. Explain what the slope of the equation means.
  2. Explain what the n–intercept of the equation means. Is this a realistic situation?

96. The equation C=5 over 9F-17.8 can be used to convert temperatures F, on the Fahrenheit scale to temperatures, C, on the Celsius scale.

  1. Explain what the slope of the equation means.
  2. Explain what the C–intercept of the equation means.
97. Why are all horizontal lines parallel? 98. Explain in your own words how to decide which method to use to graph a line.

Answers

2. slope m=4 and y-intercept (0,-2) 3. slope m=-3 and y-intercept (0,1)
6. slope m=-2 over 5 and y-intercept (0,3) 7. -9;(0,7)
10. 4;(0,-10) 11. -4;(0,8)
14. -8 over 3;(0,4) 15. 7 over 3;(0,-3)
18.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 4) and (1, 5).

19.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 3) and (1, negative 1).

22.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 3) and (1, 2).

23.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 2) and (1, negative 3).

26.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 3) and (5, negative 5).

27.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,1) and (3, negative 1).

30.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 2) and (3,2).

31.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 25) and (negative 50, 10).

34. horizontal line 35. vertical line
38. slope–intercept 39. intercepts
42. slope–intercept 43. horizontal line
46. intercepts 47. slope–intercept
50.

a) $28

b) $66.10

c) The slope, 2.54, means that Randy’s payment, P, increases by $2.54 when the number of units of water he used, w, increases by 1. The P–intercept means that if the number units of water Randy used was 0, the payment would be $28.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable w and runs from negative 2 to 20. The y-axis of the plane represents the variable P and runs from negative 1 to 100. The line begins at the point (0, 28) and goes through the point (15, 66.1).

51.

a) $15

b) $143

c) The slope, 0.32, means that the cost, C, increases by $0.32 when the number of miles driven, m, increases by 1. The C-intercept means that if Janelle drives 0 miles one day, the cost would be $15.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable m and runs from negative 1 to 500. The y-axis of the plane represents the variable C and runs from negative 1 to 200. The line begins at the point (0,15) and goes through the point (400,143).

54.

a) $750

b) $2418.60

c) The slope, 0.09, means that Patel’s salary, S, increases by $0.09 for every $1 increase in his sales. The S-intercept means that when his sales are $0, his salary is $750.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable w and runs from negative 1 to 20000. The y-axis of the plane represents the variable P and runs from negative 1 to 3000. The line begins at the point (0, 750) and goes through the point (18540, 2415).

55.

a) $2850

b) $4950

c) The slope, 42, means that the cost, C, increases by $42 for when the number of guests increases by 1. The C-intercept means that when the number of guests is 0, the cost would be $750.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable g and runs from negative 1 to 150. The y-axis of the plane represents the variable C and runs from negative 1 to 7000. The line begins at the point (0, 750) and goes through the point (100, 4950).

58. parallel 59. parallel
62. parallel 63. parallel
66. parallel 67. parallel
70. parallel 71. parallel
74. not parallel 75. not parallel
78. not parallel 79. not parallel
82. not parallel 83. perpendicular
86. perpendicular 87. not perpendicular
90. not perpendicular 91. perpendicular
94. perpendicular 95.

a) For every increase of one degree Fahrenheit, the number of chirps increases by four.

b) There would be -160 chirps when the Fahrenheit temperature is 0°. (Notice that this does not make sense; this model cannot be used for all possible temperatures.)

98. Answers will vary.
1. slope m=4 and y-intercept (0,-2) 3. slope m=-3 and y-intercept (0,1)
5. slope m=-2 over 5 and y-intercept (0,3) 7. -9;(0,7)
9. 4;(0,-10) 11. -4;(0,8)
13. -8 over 3;(0,4) 15. 7 over 3;(0,-3)
17.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 4) and (1, 5).

19.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 3) and (1, negative 1).

21.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 3) and (1, 2).

23.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 2) and (1, negative 3).

25.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 3) and (5, negative 5).

27.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0,1) and (3, negative 1).

29.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 2) and (3,2).

31.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 25) and (negative 50, 10).

33. horizontal line 35. vertical line
37. slope–intercept 39. intercepts
41. slope–intercept 43. horizontal line
45. intercepts 47. slope–intercept
49.

a) $28

b) $66.10

c) The slope, 2.54, means that Randy’s payment, P, increases by $2.54 when the number of units of water he used, w, increases by 1. The P–intercept means that if the number units of water Randy used was 0, the payment would be $28.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable w and runs from negative 2 to 20. The y-axis of the plane represents the variable P and runs from negative 1 to 100. The line begins at the point (0, 28) and goes through the point (15, 66.1).

51.

a) $15

b) $143

c) The slope, 0.32, means that the cost, C, increases by $0.32 when the number of miles driven, m, increases by 1. The C-intercept means that if Janelle drives 0 miles one day, the cost would be $15.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable m and runs from negative 1 to 500. The y-axis of the plane represents the variable C and runs from negative 1 to 200. The line begins at the point (0,15) and goes through the point (400,143).

53.

a) $750

b) $2418.60

c) The slope, 0.09, means that Patel’s salary, S, increases by $0.09 for every $1 increase in his sales. The S-intercept means that when his sales are $0, his salary is $750.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable w and runs from negative 1 to 20000. The y-axis of the plane represents the variable P and runs from negative 1 to 3000. The line begins at the point (0, 750) and goes through the point (18540, 2415).

55.

a) $2850

b) $4950

c) The slope, 42, means that the cost, C, increases by $42 for when the number of guests increases by 1. The C-intercept means that when the number of guests is 0, the cost would be $750.

d)
The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane represents the variable g and runs from negative 1 to 150. The y-axis of the plane represents the variable C and runs from negative 1 to 7000. The line begins at the point (0, 750) and goes through the point (100, 4950).

57. parallel 59. parallel
61. parallel 63. parallel
65. parallel 67. parallel
69. parallel 71. parallel
73. not parallel 75. not parallel
77. not parallel 79. not parallel
81. not parallel 83. perpendicular
85. perpendicular 87. not perpendicular
89. not perpendicular 91. perpendicular
93. perpendicular 95.

a) For every increase of one degree Fahrenheit, the number of chirps increases by four.

b) There would be -160 chirps when the Fahrenheit temperature is 0°. (Notice that this does not make sense; this model cannot be used for all possible temperatures.)

97. Answers will vary.

Attributions

This chapter has been adapted from “Use the Slope–Intercept Form of an Equation of a Line” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

37

6.6 Find the Equation of a Line

Learning Objectives

By the end of this section, you will be able to:

  • Find an equation of the line given the slope and y-intercept
  • Find an equation of the line given the slope and a point
  • Find an equation of the line given two points
  • Find an equation of a line parallel to a given line
  • Find an equation of a line perpendicular to a given line

How do online retailers know that ‘you may also like’ a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

You are at an exciting point in your mathematical journey as the mathematics you are studying has interesting applications in the real world.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. Data is collected and graphed. If the data points appear to form a straight line, an equation of that line can be used to predict the value of one variable based on the value of the other variable.

To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

Find an Equation of the Line Given the Slope and y-Intercept

We can easily determine the slope and intercept of a line if the equation was written in slope–intercept form, y=mx+b. Now, we will do the reverse—we will start with the slope and y-intercept and use them to find the equation of the line.

EXAMPLE 1

Find an equation of a line with slope -7 and y-intercept (0,-1).

Solution

Since we are given the slope and y-intercept of the line, we can substitute the needed values into the slope–intercept form, y=mx+b.

Name the slope. .
Name the y-intercept. .
Substitute the values into y=mx+b. .
.
.

TRY IT 1.1

Find an equation of a line with slope 2 over 5 and y-intercept (0,4).

Show answer

y=2 over 5x+4

TRY IT 1.2

Find an equation of a line with slope -1 and y-intercept (0,-3).

Show answer

y=-x-3

Sometimes, the slope and intercept need to be determined from the graph.

EXAMPLE  2

Find the equation of the line shown.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. A line intercepts the y-axis at (0, negative 4), passes through the plotted point (3, negative 2), and intercepts the x-axis at (4, 0).

Solution

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope–intercept form, y=mx+b.

To find the slope, we choose two points on the graph.

The y-intercept is (0,-4) and the graph passes through (3,-2).

Find the slope by counting the rise and run. .
.
Find the y-intercept. .
Substitute the values into y=mx+b. .
.

TRY IT 2.1

Find the equation of the line shown in the graph.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. A line intercepts the x-axis at (negative 2, 0), intercepts the y-axis at (0, 1) and passes through the plotted point (5, 4).

Show answer

y=3 over 5x+1

TRY IT 2.2

Find the equation of the line shown in the graph.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. A line intercepts the y-axis at (0, negative 5), passes through the plotted point (3, negative 1), and intercepts the x-axis at (15 fourths, 0).

Show answer

y=4 over 3x-5

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope–intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line. Suppose we have a line that has slope m and that contains some specific point (x sub 1,y sub 1) and some other point, which we will just call (x,y). We can write the slope of this line and then change it to a different form.

m=y-y sub 1 over x-x sub 1
Multiply both sides of the equation by x-x sub 1. mathematical expression
Simplify. mathematical expression
Rewrite the equation with the y terms on the left. mathematical expression

This format is called the point–slope form of an equation of a line.

Point–slope form of an equation of a line

The point–slope form of an equation of a line with slope m and containing the point (x sub 1,y sub 1) is

No alt text

We can use the point–slope form of an equation to find an equation of a line when we are given the slope and one point. Then we will rewrite the equation in slope–intercept form. Most applications of linear equations use the the slope–intercept form.

EXAMPLE 3

Find an Equation of a Line Given the Slope and a Point

Find an equation of a line with slope m=2 over 5 that contains the point (10,3). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Identify the slope.” The text in the second cell reads: “The slope is given.” The third cell contains the slope of a line, defined as m equals 2 fifths.In the second row, the first cell reads: “Step 2. Identify the point.” The second cell reads: “The point is given.” The third cell contains the ordered pair (10, 3). A superscript x subscript 1 is written over 10, and a superscript y subscript 1 is written over 3.In the third row, the first cell reads: “Step 3. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top line of the second cell is left blank. The third cell features the point-slope form written again: y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the point-slope form with 10 substituted for x subscript 1, 3 substituted for y subscript 1, and 2 fifths substituted for m: y minus 3 equals 2 fifths times x minus 10 in parentheses. One line down, the instructions in the second cell say: “Simplify.” In the third cell is y minus 3 equals 2 fifths x minus 4.In the fourth row, the first cell reads: “Write the equation in slope-intercept form.” The second cell is blank. In the third cell is y equals 2 fifths x minus 1.

TRY IT 3.1

Find an equation of a line with slope m=5 over 6 and containing the point (6,3).

Show answer

y=5 over 6x-2

TRY IT 3.2

Find an equation of a line with slope m=2 over 3 and containing thepoint (9,2).

Show answer

y=2 over 3x-4

HOW TO: Find an equation of a line given the slope and a point

  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
  4. Write the equation in slope–intercept form.

EXAMPLE 4

Find an equation of a line with slope m=-1 over 3 that contains the point (6,-4). Write the equation in slope–intercept form.

Solution

Since we are given a point and the slope of the line, we can substitute the needed values into the point–slope form, y-y sub 1=m(x-x sub 1).

Identify the slope. .
Identify the point. .
Substitute the values into y-y sub 1=m(x-x sub 1). .
.
Simplify. .
Write in slope–intercept form. .

TRY IT 4.1

Find an equation of a line with slope m=-2 over 5 and containing the point (10,-5).

Show answer

y=-2 over 5x-1

TRY IT 4.2

Find an equation of a line with slope m=-3 over 4, and containing the point (4,-7).

Show answer

y=-3 over 4x-4

EXAMPLE 5

Find an equation of a horizontal line that contains the point (-1,2). Write the equation in slope–intercept form.

Solution

Every horizontal line has slope 0. We can substitute the slope and points into the point–slope form, y-y sub 1=m(x-x sub 1).

Identify the slope. .
Identify the point. .
Substitute the values into y-y sub 1=m(x-x sub 1). .
.
Simplify. .
.
.
Write in slope–intercept form. It is in y-form, but could be written y=0x+2.

Did we end up with the form of a horizontal line, y=a?

TRY IT 5.1

Find an equation of a horizontal line containing the point (-3,8).

Show answer

y=8

TRY IT 5.2

Find an equation of a horizontal line containing the point (-1,4).

Show answer

y=4

Find an Equation of the Line Given Two Points

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

We have two options so far for finding an equation of a line: slope–intercept or point–slope. Since we will know two points, it will make more sense to use the point–slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

EXAMPLE 6

Find an Equation of a Line Given Two Points

Find an equation of a line that contains the points (5,4) and (3,6). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Find the slope using the given points.” The text in the second cell reads: “To use the point-slope form, we first find the slope.” The third cell contains the slope of a line formula: m equals y superscript 2 minus y superscript 1 divided by x superscript 2 minus x superscript 1. Below this is m equals 6 minus 4 divided by 3 minus 5. Below this is m equals 2 divided by negative 2. Below this is m equals negative 1.In the second row, the first cell reads: “Step 2. Choose one point.” The second cell reads: “Choose either point.” The third cell contains the ordered pair (5, 4) with a superscript x subscript 1 over 5 and a superscript y subscript 1 over 4.In the third row, the first cell reads: “Step 3. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top line of the second cell is left blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the point-slope form with 5 substituted for x subscript 1, 4 substituted for y subscript 1, and negative 1 substituted for m: y minus 4 equals negative 1 times x minus 5 in parentheses. Below this is y minus 4 equals negative x plus 5.In the fourth row, the first cell reads: “Step 4. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative x plus 9.

Use the point (3,6) and see that you get the same equation.

TRY IT 6.1

Find an equation of a line containing the points (3,1) and (5,6).

Show answer

y=5 over 2x-13 over 2

TRY IT 6.2

Find an equation of a line containing the points (1,4) and (6,2).

Show answer

y=-2 over 5x+22 over 5

HOW TO: Find an equation of a line given two points

  1. Find the slope using the given points.
  2. Choose one point.
  3. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
  4. Write the equation in slope–intercept form.

EXAMPLE 7

Find an equation of a line that contains the points (-3,-1) and (2,-2). Write the equation in slope–intercept form.

Solution

Since we have two points, we will find an equation of the line using the point–slope form. The first step will be to find the slope.

Find the slope of the line through (−3, −1) and (2, −2). .
.
.
.
Choose either point. .
Substitute the values into y-y sub 1=m(x-x sub 1). .
.
.
Write in slope–intercept form. .

TRY IT 7.1

Find an equation of a line containing the points (-2,-4) and (1,-3).

Show answer

y=1 over 3x-10 over 3

TRY IT 7.2

Find an equation of a line containing the points (-4,-3) and (1,-5).

Show answer

y=-2 over 5x-23 over 5

EXAMPLE 8

Find an equation of a line that contains the points (-2,4) and (-2,-3). Write the equation in slope–intercept form.

Solution

Again, the first step will be to find the slope.

Find the slope of the line through (-2,4) and (-2,-3). m=y sub 2-y sub 1 over x sub 2-x sub 1
m=-3-4 over -2-(-2)
m=-7 over 0
The slope is undefined.

This tells us it is a vertical line. Both of our points have an x-coordinate of -2. So our equation of the line is x=-2. Since there is no y, we cannot write it in slope–intercept form.

You may want to sketch a graph using the two given points. Does the graph agree with our conclusion that this is a vertical line?

TRY IT 8.1

Find an equation of a line containing the points (5,1) and (5,-4).

Show answer

x=5

TRY IT 8.2

Find an equation of a line containing the points (-4,4) and (-4,3).

Show answer

x=-4

We have seen that we can use either the slope–intercept form or the point–slope form to find an equation of a line. Which form we use will depend on the information we are given. This is summarized in the following table.

To Write an Equation of a Line
If given: Use: Form:
Slope and y-intercept slope–intercept y=mx+b
Slope and a point point–slope y-y sub 1=m(x-x sub 1)
Two points point–slope y-y sub 1=m(x-x sub 1)

Find an Equation of a Line Parallel to a Given Line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point–slope equation.

First let’s look at this graphically.

The graph shows the graph of y=2x-3. We want to graph a line parallel to this line and passing through the point (-2,1).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere on the graph, the point (negative 2, 1) is plotted.

We know that parallel lines have the same slope. So the second line will have the same slope asy=2x-3. That slope ismathematical expression. We’ll use the notation mathematical expression to represent the slope of a line parallel to a line with slope m. (Notice that the subscript mathematical expression looks like two parallel lines.)

The second line will pass through (-2,1) and have m=2. To graph the line, we start at(-2,1) and count out the rise and run. With m=2 (or m=2 over 1), we count out the rise 2 and the run 1. We draw the line.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). The points (negative 2, 1) and (negative 1, 3) are plotted. A second line, parallel to the first, intercepts the x-axis at (negative 5 halves, 0), passes through the points (negative 2, 1) and (negative 1, 3), and intercepts the y-axis at (0, 5).

Do the lines appear parallel? Does the second line pass through (-2,1)?

Now, let’s see how to do this algebraically.

We can use either the slope–intercept form or the point–slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point–slope form.

EXAMPLE 9

How to Find an Equation of a Line Parallel to a Given Line

Find an equation of a line parallel to y=2x-3 that contains the point (-2,1). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Find the slope of the given line.” The second cell reads: “The line is in slope-intercept form. y equals 2x minus 3.” The third cell contains the slope of a line, defined as m equals 2.In the second row, the first cell reads: “Step 2. Find the slope of the parallel line.” The second cell reads “Parallel lines have the same slope.” The third cell contains the slope of the parallel line, defined as m parallel equals 2.In the third row, the first cell reads “Step 3. Identify the point.” The second cell reads “The given point is (negative 2, 1).” The third cell contains the ordered pair (negative 2, 1) with a superscript x subscript 1 above negative 2 and a superscript y subscript 1 above 1.In the fourth row, the first cell reads “Step 4. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top of the second cell is blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the form with negative 2 substituted for x subscript 1, 1 substituted for y subscript 1, and 2 substituted for m: y minus 1 equals 2 times x minus negative 2 in parentheses. One line down, the text in the second cell says “Simplify.” The right column contains y minus 1 equals 2 times x plus 2. Below this is y minus 1 equals 2x plus 4.In the fifth row, the first cell says “Step 5. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals 2x plus 5.

Does this equation make sense? What is the y-intercept of the line? What is the slope?

TRY IT 9.1

Find an equation of a line parallel to the line y=3x+1 that contains the point (4,2). Write the equation in slope–intercept form.

Show answer

y=3x-10

TRY IT 9.2

Find an equation of a line parallel to the line y=1 over 2x-3 that contains the point (6,4).

Show answer

y=1 over 2x+1

HOW TO: Find an equation of a line parallel to a given line

  1. Find the slope of the given line.
  2. Find the slope of the parallel line.
  3. Identify the point.
  4. Substitute the values into the point–slope form, y-y sub 1=m(x-x sub 1).
  5. Write the equation in slope–intercept form.

Find an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point–slope equation, like we did with parallel lines.

The graph shows the graph of y=2x-3. Now, we want to graph a line perpendicular to this line and passing through (-2,1).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere on the graph, the point (negative 2, 1) is plotted.

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation mathematical expression to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

mathematical expression

We now know the perpendicular line will pass through (-2,1) with mathematical expression.

To graph the line, we will start at (-2,1) and count out the rise -1 and the run 2. Then we draw the line.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere, the point (negative 2, 1) is plotted. Another line perpendicular to the first line passes through the point (negative 2, 1) and intercepts the x and y-axes at (0, 0). A red line with an arrow extends left from (0, 0) to (negative 2, 0), then extends up and terminates at (negative 2, 1), forming a right triangle with the second line as a hypotenuse.

Do the lines appear perpendicular? Does the second line pass through (-2,1)?

Now, let’s see how to do this algebraically. We can use either the slope–intercept form or the point–slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point–slope form.

EXAMPLE 10

How to Find an Equation of a Line Perpendicular to a Given Line

Find an equation of a line perpendicular to y=2x-3 that contains the point (-2,1). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Find the slope of the given line.” The second cell reads: “The line is in slope-intercept form. y equals 2x minus 3.” The third cell contains the slope of a line, defined as m equals 2.In the second row, the first cell reads: “Step 2. Find the slope of the perpendicular line.” The second cell reads “The slopes of perpendicular lines are negative reciprocals.” The third cell contains the slope of the perpendicular line, defined as m perpendicular equals negative 1 half.In the third row, the first cell reads “Step 3. Identify the point.” The second cell reads “The given point is (negative 2, 1).” The third cell contains the ordered pair (negative 2, 1) with a superscript x subscript 1 above negative 2 and a superscript y subscript 1 above 1.In the fourth row, the first cell reads “Step 4. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top of the second cell is blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the form with negative 2 substituted for x subscript 1, 1 substituted for y subscript 1, and negative 1 half substituted for m: y minus 1 equals negative 1 half times x minus negative 2 in parentheses. One line down, the text in the second cell says “Simplify.” The right column contains y minus 1 equals negative 1 half times x plus 2. Below this is y minus 1 equals negative 1 half x plus minus 1.In the fifth row, the first cell says “Step 5. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative 1 half x.

TRY IT 10.1

Find an equation of a line perpendicular to the line y=3x+1 that contains the point (4,2). Write the equation in slope–intercept form.

Show answer

y=-1 over 3x+10 over 3

TRY IT 10.2

Find an equation of a line perpendicular to the line y=1 over 2x-3 that contains the point (6,4).

Show answer

y=-2x+16

HOW TO: Find an equation of a line perpendicular to a given line

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point–slope form, y-y sub 1=m(x-x sub 1).
  5. Write the equation in slope–intercept form.

EXAMPLE 11

Find an equation of a line perpendicular to x=5 that contains the point (3,-2). Write the equation in slope–intercept form.

Solution

Again, since we know one point, the point–slope option seems more promising than the slope–intercept option. We need the slope to use this form, and we know the new line will be perpendicular to x=5. This line is vertical, so its perpendicular will be horizontal. This tells us the mathematical expression.

Identify the point. (3,-2)
Identify the slope of the perpendicular line. mathematical expression
Substitute the values into y-y sub 1=m(x-x sub 1). mathematical expression
Simplify. y=-2

Sketch the graph of both lines. Do they appear to be perpendicular?

TRY IT 11.1

Find an equation of a line that is perpendicular to the line x=4 that contains the point (4,-5). Write the equation in slope–intercept form.

Show answer

y=-5

TRY IT 11.2

Find an equation of a line that is perpendicular to the line x=2 that contains the point (2,-1). Write the equation in slope–intercept form.

Show answer

y=-1

In (Example 11), we used the point–slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to x=5 that contains the point (3,-2). The graph shows us the linex=5 and the point (3,-2).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is x equals 5 intercepts the x-axis at (5, 0) and runs parallel to the y-axis. Elsewhere on the graph, the point (3, negative 2) is plotted.

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through (3,-2).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is x equals 5 intercepts the x-axis at (5, 0) and runs parallel to the y-axis. Elsewhere on the graph, the points (negative 2, negative 2), (0, negative 2), (3, negative 2), and (6, negative 2) are plotted. A line perpendicular to the previous line passes through those points and runs parallel to the x-axis.

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of -2. So, the equation of the line perpendicular to the vertical line x=5 is y=-2.

EXAMPLE 12

Find an equation of a line that is perpendicular to y=-4 that contains the point (-4,2).

Write the equation in slope–intercept form.

Solution

The line y=-4 is a horizontal line. Any line perpendicular to it must be vertical, in the form x=a. Since the perpendicular line is vertical and passes through (-4,2), every point on it has an x-coordinate of -4. The equation of the perpendicular line is x=-4. You may want to sketch the lines. Do they appear perpendicular?

TRY IT 12.1

Find an equation of a line that is perpendicular to the line y=1 that contains the point (-5,1). Write the equation in slope–intercept form.

Show answer

x=-5

TRY IT 12.1

Find an equation of a line that is perpendicular to the line y=-5 that contains the point (-4,-5).

Show answer

x=-4

Access this online resource for additional instruction and practice with finding the equation of a line.

Key Concepts

  • To Find an Equation of a Line Given the Slope and a Point
    1. Identify the slope.
    2. Identify the point.
    3. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
    4. Write the equation in slope-intercept form.
  • To Find an Equation of a Line Given Two Points
    1. Find the slope using the given points.
    2. Choose one point.
    3. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
    4. Write the equation in slope-intercept form.
  • To Write and Equation of a Line
    • If given slope and y-intercept, use slope–intercept form y=mx+b.
    • If given slope and a point, use point–slope form y-y sub 1=m(x-x sub 1).
    • If given two points, use point–slope form y-y sub 1=m(x-x sub 1).
  • To Find an Equation of a Line Parallel to a Given Line
    1. Find the slope of the given line.
    2. Find the slope of the parallel line.
    3. Identify the point.
    4. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
    5. Write the equation in slope-intercept form.
  • To Find an Equation of a Line Perpendicular to a Given Line
    1. Find the slope of the given line.
    2. Find the slope of the perpendicular line.
    3. Identify the point.
    4. Substitute the values into the point-slope form, y-y sub 1=m(x-x sub 1).
    5. Write the equation in slope-intercept form.

Glossary

point–slope form
The point–slope form of an equation of a line with slope m and containing the point (x sub 1,y sub 1) is y-y sub 1=m(x-x sub 1).

Practice Makes Perfect

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

1. slope 4 and y-intercept (0,1) 2. slope 3 and y-intercept (0,5)
3. slope 8 and y-intercept (0,-6) 4. slope 6 and y-intercept (0,-4)
5. slope -1 and y-intercept (0,7) 6. slope -1 and y-intercept (0,3)
7. slope -3 and y-intercept (0,-1) 8. slope -2 and y-intercept (0,-3)
9. slope 1 over 5 and y-intercept (0,-5) 10. slope 3 over 5 and y-intercept (0,-1)
11. slope -2 over 3 and y-intercept (0,-3) 12. slope -3 over 4 and y-intercept (0,-2)
13. slope 0 and y-intercept (0,2) 14. slope 0 and y-intercept (0,-1)
15. slope -4 and y-intercept (0,0) 16. slope -3 and y-intercept (0,0)

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

17. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, 0) is plotted. A line intercepts the y-axis at (0, 4) and intercepts the x-axis at (2, 0). 18. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (1, negative 2) is plotted. A line intercepts the y-axis at (0, negative 5), passes through the point (1, negative 2), and intercepts the x-axis at (5 thirds, 0).
19. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (4, 5) is plotted. A line intercepts the x-axis at (negative 8 thirds, 0), intercepts the y-axis at (0, 2), and passes through the point (4, 5). 20. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (6, 0) is plotted. A line intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (6, 0).
21. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, negative 4) is plotted. A line intercepts the x-axis at (negative 2 thirds, 0), intercepts the y-axis at (0, negative 1), and passes through the point (2, negative 4). 22. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (3, negative 1) is plotted. A line intercepts the y-axis at (0, 2), intercepts the x-axis at (9 fourths, 0), and passes through the point (3, negative 1).
23. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (negative 3, 6) is plotted. A line running parallel to the x-axis passes through (negative 3, 6) and intercepts the y-axis at (0, 6). 24. The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, negative 2) is plotted. A line running parallel to the x-axis intercepts the y-axis at (0, negative 2) and passes through the point (2, negative 2).

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

25. m=3 over 8, point (8,2) 26. m=5 over 8, point (8,3)
27. m=5 over 6, point (6,7) 28. m=1 over 6, point (6,1)
29. m=-3 over 5, point (10,-5) 30. m=-3 over 4, point (8,-5)
31. m=-1 over 3, point (-9,-8) 32. m=-1 over 4, point (-12,-6)
33. Horizontal line containing (-1,4) 34. Horizontal line containing (-2,5)
35. Horizontal line containing (-1,-7) 36. Horizontal line containing (-2,-3)
37. m=-5 over 2, point (-8,-2) 38. m=-3 over 2, point (-4,-3)
39. m=-4, point (-2,-3) 40. m=-7, point (-1,-3)
41. Horizontal line containing (4,-8) 42. Horizontal line containing (2,-3)

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

43. (3,1) and (2,5) 44. (2,6) and (5,3)
45. (2,7) and (3,8) 46. (4,3) and (8,1)
47. (-5,-3) and (4,-6) 48. (-3,-4) and (5-2)
49. (-2,8) and (-4,-6) 50. (-1,3) and (-6,-7)
51. (3,-2) and (-4,4) 52. (6,-4) and (-2,5)
53. (0,-2) and (-5,-3) 54. (0,4) and (2,-3)
55. (4,2) and (4,-3) 56. (7,2) and (7,-2)
57. (-2,1) and (-2,-4) 58. (-7,-1) and (-7,-4)
59. (6,2) and (-3,2) 60. (6,1) and (0,1)
61. (-6,-3) and (-1,-3) 62. (3,-4) and (5,-4)
63. (0,0) and (1,4) 64. (4,3) and (8,0)
65. (-3,0) and (-7,-2) 66. (-2,-3) and (-5,-6)
67. (3,5) and (-7,5) 68. (8,-1) and (8,-5)

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

69. line y=3x+4, point (2,5) 70. line y=4x+2, point (1,2)
71. line y=-3x-1, point (2,-3) 72. line y=-2x-3, point (-1,3)
73. line 2x-y=6, point (3,0) 74. line 3x-y=4, point (3,1)
75. line 2x+3y=6, point (0,5) 76. line 4x+3y=6, point (0,-3)
77. line x=-4, point (-3,-5) 78. line x=-3, point (-2,-1)
79. line x-6=0, point (4,-3) 80. line x-2=0, point (1,-2)
81. line y=1, point (3,-4) 82. line y=5, point (2,-2)
83. line y+7=0, point (1,-1) 84. line y+2=0, point (3,-3)

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

85. line y=-x+5, point (3,3) 86. line y=-2x+3, point (2,2)
87. line y=2 over 3x-4, point (2,-4) 88. line y=3 over 4x-2, point (-3,4)
89. line 4x-3y=5, point (-3,2) 90. line 2x-3y=8, point (4,-1)
91. line 4x+5y=-3, point (0,0) 92. line 2x+5y=6, point (0,0)
93. line y-6=0, point (-5,-3) 94. line y-3=0, point (-2,-4)
95. line y-axis, point (2,1) 96. line y-axis, point (3,4)

Mixed Practice

In the following exercises, find the equation of each line. Write the equation in slope–intercept form.

97. Containing the points (2,7) and (3,8) 98. Containing the points (4,3) and (8,1)
99. m=5 over 6, containing point (6,7) 100. m=1 over 6, containing point (6,1)
101. Parallel to the line 2x+3y=6, containing point (0,5) 102. Parallel to the line 4x+3y=6, containing point (0,-3)
103. m=-3 over 5, containing point (10,-5) 104. m=-3 over 4, containing point (8,-5)
105. Perpendicular to the line y-axis, point (-6,2) 106. Perpendicular to the line y-1=0, point (-2,6)
107. Containing the points (-2,0) and (-3,-2) 108. Containing the points (4,3) and (8,1)
109. Parallel to the line x=-4, containing point (-3,-5) 110. Parallel to the line x=-3, containing point (-2,-1)
111. Containing the points (-5,-3) and (4,-6) 112. Containing the points (-3,-4) and (2,-5)
113. Perpendicular to the line 4x+3y=1, containing point (0,0) 114. Perpendicular to the line x-2y=5, containing point (-2,2)

Everyday Math

115. Fuel consumption. The city mpg, x, and highway mpg, y, of two cars are given by the points (29,40) and(19,28). Find a linear equation that models the relationship between city mpg and highway mpg. 116. Cholesterol. The age, x, and LDL cholesterol level, y, of two men are given by the points (18,68) and (27,122). Find a linear equation that models the relationship between age and LDL cholesterol level.

Writing Exercises

117. Explain in your own words why the slopes of two perpendicular lines must have opposite signs. 118. Why are all horizontal lines parallel?

Answers

1. y=4x+1 3. y=8x-6
5. y=-x+7 7. y=-3x-1
9. y=1 over 5x-5 11. y=-2 over 3x-3
13. y=2 15. y=-4x
17. y=-2x+4 19. y=3 over 4x+2
21. y=-3 over 2x-1 23. y=6
25. y=3 over 8x-1 27. y=5 over 6x+2
29. y=-3 over 5x+1 31. y=-1 over 3x-11
33. y=4 35. y=-7
37. y=-5 over 2x-22 39. y=-4x-11
41. y=-8 43. y=-4x+13
45. y=x+5 47. y=-1 over 3x-14 over 3
49. y=7x+22 51. y=-6 over 7x+4 over 7
53. y=1 over 5x-2 55. x=4
57. x=-2 59. y=2
61. y=-3 63. y=4x
65. y=1 over 2x+3 over 2 67. y=5
69. y=3x-1 71. y=-3x+3
73. y=2x-6 75. y=-2 over 3x+5
77. x=-3 79. x=4
81. y=-4 83. y=-1
85. y=x 87. y=-3 over 2x-1
89. y=-3 over 4x-1 over 4 91. y=5 over 4x
93. x=-5 95. y=1
97. y=x+5 99. y=5 over 6x+2
101. y=-2 over 3x+5 103. y=-3 over 5x+1
105. y=2 107. y=x+2
109. x=-3 111. y=-1 over 3x-14 over 3
113. y=3 over 4x 115. y=1.2x+5.2
117. Answers will vary.

Attributions

This chapter has been adapted from “Find the Equation of a Line” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

38

6.7 Chapter Review

Review Exercises

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.

1.
a) (4,3)
b) (-4,3)
c) (-4,-3)
d) (4,-3)
2.
a) (-1,-5)
b) (-3,4)
c) (2,-3)
d) (1,5 over 2)
3.
a) (2,3 over 2)
b) (3,4 over 3)
c) (1 over 3,-4)
d) (1 over 2,-5)
4.
a) (-2,0)
b) (0,-4)
c) (0,5)
d) (3,0)

Identify Points on a Graph

In the following exercises, name the ordered pair of each point shown in the rectangular coordinate system.

5.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (2, 0) is plotted and labeled "a". The point (0, negative 5) is plotted and labeled "b". The point (negative 4, 0) is plotted and labeled "c". The point (0, 3) is plotted and labeled “d”. 6.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (5, 3) is plotted and labeled "a". The point (2, negative 1) is plotted and labeled "b". The point (negative 3, negative 2) is plotted and labeled "c". The point (negative 1, 4) is plotted and labeled “d”.

Verify Solutions to an Equation in Two Variables

In the following exercises, which ordered pairs are solutions to the given equations?

7. y=6x-2

a) (1,4)
b) (1 over 3,0)
c) (6,-2)

8. 5x+y=10

a) (5,1)
b) (2,0)
c) (4,-10)

Complete a Table of Solutions to a Linear Equation in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

9. y=-1 over 2x+3

x y (x,y)
0
4
-2

10. y=4x-1

x y (x,y)
0
1
-2

11. 3x+2y=6

x y (x,y)
0
0
-2

12. x+2y=5

x y (x,y)
0
1
-1

Find Solutions to a Linear Equation in Two Variables

In the following exercises, find three solutions to each linear equation.

13. x+y=-4 14. x+y=3
15. y=-x-1 16. y=3x+1

Recognize the Relation Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

a) Is the ordered pair a solution to the equation?

b) Is the point on the line?

17.

y=2 over 3x-1

(0,-1) (3, 1)

(-3,-3) (6, 4)

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals two-thirds x minus 1 is plotted as an arrow extending from the bottom left toward the top right.

18.

y=-x+4

(0,4)(-1,3)

(2,2)(-2,6)

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x plus 4 is plotted as an arrow extending from the top left toward the bottom right.

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

19. y=-3x 20. y=4x-3
21. x-y=6 22. y=1 over 2x+3
23. 3x-2y=6 24. 2x+y=7

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

25. x=3 26. y=-2

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

27. y=4 over 3x and y=4 over 3 28. y=-2x and y=-2

Identify the x– and y-Intercepts on a Graph

In the following exercises, find the x– and y-intercepts.

29.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (3, 0) and (0, 3) is plotted. 30. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 4, 0) and (0, 4) is plotted.

Find the x– and y-Intercepts from an Equation of a Line

In the following exercises, find the intercepts of each equation.

31. x-y=-1 32. x+y=5
33. 2x+3y=12 34. x+2y=6
35. y=3x 36. y=3 over 4x-12

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

37. -x+3y=3 38. x+y=-2
39. x-y=4 40. 2x-y=5
41. 2x-4y=8 42. y=2x

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

43. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 4 row 1. 44. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 4 row 2.
45. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4. 46. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 3 and the point in column 4 row 4.

In the following exercises, model each slope. Draw a picture to show your results.

47. 1 over 3 48. 3 over 2
49. -2 over 3 50. -1 over 2

In the following exercises, find the slope of each line shown. Use m=rise over run to find the slope of a line from its graph.

51. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 4, 0) and (0, 4) is plotted. 52. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 1, 3), (0, 0), and (1, negative 3) is plotted.
53. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 3, 6) and (5, 2) is plotted. 54.

Find the Slope of Horizontal and Vertical Lines

In the following exercises, find the slope of each line.

55. x=5 56. y=2
57. y=-1 58. x=-3

Use the Slope Formula to find the Slope of a Line between Two Points

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

59. (3,5),(4,-1) 60. (-1,-1),(0,5)
61. (2,1),(4,6) 62. (-5,-2),(3,2)

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

63. (-3,4); m=-1 over 3 64. (2,-2); m=5 over 2
65. y-intercept 1; m=-3 over 4 66. x-intercept -4; m=3

Solve Slope Applications

In the following exercises, solve these slope applications.

67. A mountain road rises 50 feet for a 500-foot run. What is its slope? 68. The roof pictured below has a rise of 10 feet and a run of 15 feet. What is its slope?
The figure shows a person on a ladder using a hammer on the roof of a building.

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

In the following exercises, use the graph to find the slope and y-intercept of each line. Compare the values to the equation y=mx+b.

69.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals two-thirds x plus 4 is plotted from the top left to the bottom right.

y=-2 over 3x+4

70.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 4 x minus 1 is plotted from the lower left to the top right.

y=4x-1

Identify the Slope and y-Intercept from an Equation of a Line

In the following exercises, identify the slope and y-intercept of each line.

71. y=5 over 3x-6 72. y=-4x+9
73. 4x-5y=8 74. 5x+y=10

Graph a Line Using Its Slope and Intercept

In the following exercises, graph the line of each equation using its slope and y-intercept.

75. y=-x-1 76. y=2x+3
77. 4x-3y=12 78. y=-2 over 5x+3

In the following exercises, determine the most convenient method to graph each line.

79. y=-3 80. x=5
81. x-y=2 82. 2x+y=5
83. y=3 over 4x-1 84. y=x+2

Graph and Interpret Applications of Slope–Intercept

85. Marjorie teaches piano. The equation P=35h-250 models the relation between her weekly profit, P, in dollars and the number of student lessons, s, that she teaches.

  1. Find Marjorie’s profit for a week when she teaches no student lessons.
  2. Find the profit for a week when she teaches 20 student lessons.
  3. Interpret the slope and P–intercept of the equation.
  4. Graph the equation.

86. Katherine is a private chef. The equation C=6.5m+42 models the relation between her weekly cost, C, in dollars and the number of meals, m, that she serves.

  1. Find Katherine’s cost for a week when she serves no meals.
  2. Find the cost for a week when she serves 14 meals.
  3. Interpret the slope and C-intercept of the equation.
  4. Graph the equation.

Use Slopes to Identify Parallel Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel.

87. mathematical expression 88. mathematical expression

Use Slopes to Identify Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are perpendicular.

89. 3x-2y=5;2x+3y=6 90. y=5x-1;10x+2y=0

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

91. slope -5 and y-intercept (0,-3) 92. slope 1 over 3 and y-intercept (0,-6)
93. slope -2 and y-intercept (0,0) 94. slope 0 and y-intercept (0,4)

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

95. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 3 x plus 5 is plotted from the top left to the bottom right. 96. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 x plus 1 is plotted from the bottom left to the top right.
97. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 4 is plotted as a horizontal line. 98. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals three-fourths x minus 2 is plotted from the bottom left to the top right.

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

99. m=3 over 5, point (10,6) 100. m=-1 over 4, point (-8,3)
101. m=-2, point (-1,-3) 102. Horizontal line containing (-2,7)

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

103. (7,1) and (5,0) 104. (2,10) and (-2,-2)
105. (5,2) and (-1,2) 106. (3,8) and (3,-4).

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

107. line 2x+5y=-10, point (10,4) 108. line y=-3x+6, point (1,-5)
109. line y=-5, point (-4,3) 110. line x=4, point (-2,-1)

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

112. line 2x-3y=9, point (-4,0) 111. line y=-4 over 5x+2, point (8,9)
114. line x=-5 point (2,1) 113. line y=3, point (-1,-3)

Review Answers

1. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (4, 3) is plotted and labeled "a". The point (negative 4, 3) is plotted and labeled "b". The point (negative 4, negative 3) is plotted and labeled "c". The point (4, negative 3) is plotted and labeled “d”. 3. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (2, three halves) is plotted and labeled "a". The point (3, four thirds) is plotted and labeled "b". The point (one third, negative 4) is plotted and labeled "c". The point (one-half, negative 5) is plotted and labeled “d”.
5. a) (2,0) b) (0,-5) c) (-4.0) d) (0,3) 7. a, b
9.
x y (x,y)
0 3 (0,3)
4 1 (4, 1)
-2 4 (-2,4)
11.
x y (x,y)
0 -3 (0,-3)
2 0 (2,0)
-2 -6 (-2,-6)
13. Answers will vary. 15. Answers will vary.
17. a) yes; yes b) yes; no 19.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 3 x is plotted as an arrow extending from the top left toward the bottom right.
21. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x minus y equals 6 is plotted as an arrow extending from the bottom left toward the top right. 23. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 3 x minus 2 y equals 6 is plotted as an arrow extending from the bottom left toward the top right.
25. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x equals 3 is plotted as a vertical line. 27. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals four-thirds x is plotted as an arrow extending from the bottom left toward the top right. The line y equals four-thirds is plotted as a horizontal line.
29. (3,0),(0,3) 31. (-1,0),(0,1)
33. (6,0),(0,4) 35. (0,0)
37. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x plus y equals negative 2 is plotted as an arrow extending from the top left toward the bottom right. 39. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 2 x minus y equals 5 is plotted as an arrow extending from the bottom left toward the top right.
41. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 x is plotted as an arrow extending from the bottom left toward the top right. 43. 4 over 3
45. -2 over 3 47. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 3 row 2.
49. The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 2 and the point in column 3 row 3. 51. 1
53. -1 over 2 55. undefined
57. 0 59. -6
61. 5 over 2 63. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 3, 4) and (0, 3) is plotted.
65. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (0, 1) and (4, negative 2) is plotted. 67. 1 over 10
69. slope m=-2 over 3 and y-intercept (0,4) 71. 5 over 3;(0,-6)
73. 4 over 5;(0,-8 over 5) 75. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x minus 1 is plotted from the top left to the bottom right.
77. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 4 x minus 3 y equals 12 is plotted from the bottom left to the top right. 79. horizontal line
81. intercepts 83. plotting points
85. a) −?250 b) ?450 c) The slope, 35, means that Marjorie’s weekly profit, P, increases by $35 for each additional student lesson she teaches. The P–intercept means that when the number of lessons is 0, Marjorie loses $250. d)
The graph shows the x y-coordinate plane where h is plotted along the x-axis and P is potted along the y-axis. The x-axis runs from 0 to 24. The y-axis runs from negative 300 to 500. The line P equals 35 h minus 250 is plotted from the bottom left to the top right.
87. not parallel
89. perpendicular 91. y=-5x-3
93. y=-2x 95. y=-3x+5
97. y=-4 99. y=3 over 5x
101. y=-2x-5 103. y=1 over 2x-5 over 2
105. y=2 107. y=-2 over 5x+8
109. y=3 111. y=-3 over 2x-6
113. y=1

Practice Test

1. Plot each point in a rectangular coordinate system.

a) (2,5)
b) (-1,-3)
c) (0,2)
d) (-4,3 over 2)
e) (5,0)

2. Which of the given ordered pairs are solutions to the equation 3x-y=6?

a) (3,3)
b) (2,0)
c) (4,-6)

3. Find three solutions to the linear equation y=-2x-4. 4. Find the x– and y-intercepts of the equation 4x-3y=12.

Find the slope of each line shown.

5. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 5, 2) and (0, negative 1) is plotted from the top left toward the bottom right. 6. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A vertical line passing through the point (2, 0) is plotted.
7. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A horizontal line passing through the point (0, 5) is plotted.
8. Find the slope of the line between the points (5,2) and (-1,-4). 9. Graph the line with slope 1 over 2 containing the point (-3,-4).

Graph the line for each of the following equations

10. y=5 over 3x-1 11. y=-x
12. x-y=2 13. 4x+2y=-8
14. y=2 15. x=-3

Find the equation of each line. Write the equation in slope–intercept form.

16. slope -3 over 4 and y-intercept (0,-2) 17. m=2, point (-3,-1)
18. containing (10,1) and (6,-1) 19. parallel to the line y=-2 over 3x-1, containing the point (-3,8)
20. perpendicular to the line y=5 over 4x+2, containing the point (-10,3)

Practice Test Answers

1. 2. a) yes b) yes c) no
3. Answer may vary 4. (3,0),(0,-4)
5. m = frac35 6. undefined
7. m = 0 8. 1
9. y = 1 over 2x – 5 over 2 10. The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals five-thirds x minus 1 is plotted. The line passes through the points (0, negative 1) and (three-fifths, 0).
11.  12.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x minus y equals 2 is plotted. The line passes through the points (0, negative 2) and (2, 0).
13. 14.The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 is plotted as a horizontal line passing through the point (0, 2).
15. 16. y=-3 over 4x-2
17.y = 2x + 5 18. y=1 over 2x-4
19. y = -2 over 3x+6 20. y=-4 over 5x-5

Attributions

This chapter has been adapted from “Review Exercises” and “Practice Test” in Chapter 4 of Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.